\section{Introduction}

In this paper we investigate the automorphism group of the
superextension $\lambda(S)$ of a  finite monogenic semigroup $S$. The
thorough study of various extensions of semigroups was started in
\cite{G2} and continued in \cite{BG2}--\cite{BGN} and \cite{G4}--\cite{G8}.
The largest among these extensions is the semigroup $\upsilon(S)$ of all
upfamilies on $S$. A family $\mathcal{A}$ of non-empty subsets of
a set $X$ is called an {\em upfamily} if for each set
$A\in\mathcal{A}$ any subset $B\supset A$ of $X$ belongs to
$\mathcal{A}$. Each family $\mathcal{B}$ of non-empty subsets of
$X$ generates the upfamily $\langle B \subset X :
B\in\mathcal{B}\rangle = \{ A \subset X : \exists B \in\mathcal{B}\
( B \subset A )\}$. An upfamily $\mathcal{F}$  that is closed
under taking  finite intersections is called a {\em filter}. A
filter $\mathcal{U}$ is called an {\em ultrafilter} if
$\mathcal{U} = \mathcal{F}$ for any filter $\mathcal{F}$
containing $\mathcal{U}$. The family $\beta(X)$ of all
ultrafilters on a set $X$ is called {\em the Stone-\v Cech
extension of $X$}, see \cite{HS}. An ultrafilter, generated by a singleton $\{ x \}$, $x\in X$, is called
{\em principal}. Each point $x\in X$ is identified with the
principal ultrafilter $\langle\{ x \}\rangle$ generated by the
singleton $\{ x \}$, and hence we can consider
$X\subset\beta(X)\subset\upsilon(X)$. It was shown in \cite{G2}
that any associative binary operation $* : S\times S \to S$ can be
extended to an associative binary operation $*:
\upsilon(S)\times\upsilon(S)\to\upsilon(S)$ by the formula
\begin{equation*}
\mathcal A*\mathcal B=\big\langle\bigcup_{a\in A}a*B_a:A\in\mathcal A,\;\;\{B_a\}_{a\in A}\subset\mathcal B\big\rangle
\end{equation*}
for upfamilies $\mathcal{A}, \mathcal{B}\in\upsilon(S)$. In this
case the Stone-\v Cech compactification $\beta(S)$ is a
subsemigroup of the semigroup $\upsilon(S)$.



The semigroup $\upsilon(S)$ contains as subsemigroups many other important
extensions of $S$. In particular, it contains the semigroup
$\lambda(S)$ of maximal linked upfamilies, see \cite{G1}, \cite{G2}.
An upfamily $\mathcal L$ of
subsets  of $S$ is said to be {\em linked} if $A\cap
B\ne\emptyset$ for all $A,B\in\mathcal L$.  A linked upfamily
$\mathcal M$ of subsets of $S$ is  {\em maximal linked} if
$\mathcal M$ coincides with each linked upfamily $\mathcal L$ on
$S$ that contains $\mathcal M$. It follows that $\beta(S)$ is a subsemigroup of $\lambda(S)$.
The space $\lambda(S)$
is well-known in  General and Categorial Topology as the {\em
superextension} of $S$, see \cite{vM}--\cite{Ve}.



For a finite set $X$, the cardinality of the set $\lambda(X)$ grows very quickly as $|X|$ tends to infinity.
The calculation of the cardinality of $\lambda(X)$ seems to be a
difficult combinatorial problem, which can be reformulated as the problem of counting the number $\lambda(n)$ of  self-dual monotone Boolean functions of $n$ variables, which is well-known in Discrete Mathematics.
According to Proposition 1.1 in \cite{BMMV}, $$\log_2\lambda(n)=\frac{2^n}{\sqrt{2\pi n}}+o(1),$$ which means that the sequence $(\lambda(n))_{n=1}^\infty$ has double exponential growth.
The sequence of numbers  $\lambda(n)$ (known in Discrete Mathematics as Ho\c sten-Morris numbers) is included in the {\tt On-line Encyclopedia of Integer Sequences} as the sequence A001206. All known precise values of this sequence (taken from \cite{BMMV}) are presented the following table.

\begin{center}
\begin{tabular}{|c|ccccccccc|}
\hline
$|X|$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 &9\\
\hline
$|\lambda(X)|$ & 1 & 2 & 4 & 12 & 81 & 2646 & 1422564 &229809982112&423295099074735261880\\
\hline
\end{tabular}
\end{center}
\bigskip

 Each map $f:X\to Y$ induces the map
$$\lambda f:\lambda (X)\to \lambda (Y),\quad \lambda f:\mathcal L\mapsto
\big\langle f(L)\subset Y: L\in\mathcal L\big\rangle, \ \text{ see \cite{G1}}.$$

If $\varphi: S\to S'$ is a homomorphism of semigroups, then $\lambda \varphi:
\lambda(S)\to \lambda(S')$ is a homomorphism as well, see \cite{G4}.


A non-empty subset $I$ of a semigroup $S$ is called an {\em  ideal}
if $IS\cup SI\subset I$. An ideal $I$ of a semigroup $S$ is called  {\em proper} if $I\neq S$.
A proper ideal $M$ of $S$ is  {\em maximal} if $M$ coincides with each proper ideal $I$ of $S$ that contains $M$.
It is easy to see that for every $n\in \mathbb N$ the subset $S^{\cdot n}=\{x_1\cdot\ldots\cdot x_n:x_1,\dots,x_n\in S\}$ is an ideal in $S$.

An element $z$ of a semigroup $S$ is called a {\em zero} (resp. a {\em  left zero}, a {\em right
zero}) in $S$ if $az=za=z$ (resp. $za=z$, $az=z$) for any $a\in
S$. An element $e$ of a semigroup $S$ is called an {\em idempotent} if $ee=e$. By $E(S)$ we denote the
set of all idempotents of a semigroup $S$.


Recall that an {\em isomorphism} between $S$ and $S'$ is a bijective function
$\psi:S\to S'$ such that $\psi(xy)=\psi(x)\psi(y)$ for
all $x,y\in S$. If there exists an isomorphism between $S$ and
$S'$, then $S$ and $S'$ are said to be {\em isomorphic}, denoted
$S\cong S'$. An isomorphism $\psi:S\to S$ is called an {\em automorphism} of a semigroup $S$. By $\mathrm{Aut\mkern 2mu}(S)$ we denote
the automorphism group of a semigroup $S$.


For an automorphism $\psi$ of a semigroup $S$, a subset $A\subset S$ is called {\em $\psi$-invariant} if $\psi(A)=A$. A subset $A\subset S$ is called {\em characteristic} if $\psi(A)=A$ for any automorphism $\psi$ of $S$. It is easy to see that the set $E(S)$ is characteristic in $S$ and so are the ideals $S^{{\cdot} n}$ for all $n\in\mathbb N$.


For a set $X$ by $S_X$ we denote the group of all bijections of $X$. For two sets $X\subset Y$ we shall identify $S_X$ with the subgroup $\{\varphi\in S_Y:\varphi|Y\setminus X=\mathrm{id}\}$ of the permutation group $S_Y$. By $C_n$ we denote a cyclic group of order $n\in\mathbb N$.


In this paper we study automorphisms of superextensions of finite monogenic semigroups.
The thorough study of automorphism groups of  superextensions of semigroups was started in
\cite{G9} and continued in \cite{BG9}.
In \cite{G9} it was shown that each automorphism of a semigroup $S$ can be extended to an automorphism of its superextension $\lambda(S)$
and the automorphism group $\mathrm{Aut\mkern 2mu}(\lambda(S))$ of the superextension of a semigroup $S$ contains a subgroup,
isomorphic to the auto\-morphism group $\mathrm{Aut\mkern 2mu}(S)$ of $S$. Also  the automorphism groups of  superextensions
of null semigroups, almost null semigroups, right zero semigroups,  left zero semigroups and all  three-element semigroups were  described.
In \cite{BG9} we studied the automorphism groups  of super\-exten\-sions of groups and described  the automorphism groups $\mathrm{Aut\mkern 2mu}(\lambda(G))$
of the super\-exten\-sions of all groups $G$ of cardinality $|G|\leq 5$.
The obtained results  are presented in Table~\ref{tab:dedekind_gr}.

\begin{table}[ht]
\centering
\begin{tabular}{|r|cccccc|}
\hline
$G\cong$ & $C_1$ & $C_2$ & $C_3$ & $C_4$ & $C_2\times C_2$&$C_5$\\
\hline
$\mathrm{Aut\mkern 2mu}(G)\cong$ & $C_1$& $C_1$ & $C_2$ & $C_2$ & $S_3$ &$C_4$\\
\hline
$\mathrm{Aut\mkern 2mu}(\lambda(G))\cong$ & $C_1$ & $C_1$ & $C_2$ & $C_2\times C_2$ & $S_4$ &$C_4$\\
\hline
\end{tabular}
\smallskip
\caption{The automorphism groups of superextensions of groups of cardinality $\le 5$.}\label{tab:dedekind_gr}
\end{table}
\vskip-18pt

In Section~\ref{monsem} we establish some general results on the structure of superextensions of finite monogenic groups, using the notion of a good shift introduced in Section~\ref{s:good}. In particular, in Theorem~\ref{t3.2} we prove that two monogenic groups are isomorphic if and only if their superextensions are isomorphic. In Section~\ref{s4} we use the results of Section~\ref{monsem} to classify the authomorphism groups of the superextensions of monogenic semigroups of order $\le 5$. The obtained results are summed up in the table in Section~\ref{s:table}. 





\section{Semigroups possessing a good shift}\label{s:good}

In this section we develop a tool for recognizing the automorphism group of a semigroup possessing a good shift.

\begin{definition}\label{d:good} Let $X$ be a semigroup. A function $\sigma:X\to X$ is called a {\em good shift} of $X$ if
$X\cdot X\subset \sigma(X)$ and for any elements $x,y\in X$ with $\sigma(x)=\sigma(y)$ the equalities $xz=yz$ and $zx=zy$ hold for all $z\in X$.

A good shift $\sigma:S\to S$ is called an {\em auto-shift} if $\sigma\circ\psi=\psi\circ\sigma$ for any automorphism $\psi\in\mathrm{Aut\mkern 2mu}(S)$.

\end{definition}



\begin{proposition}\label{p:kernel} If $\sigma:X\to X$ is a good shift of a semigroup $X$, then
the automorphism group $\mathrm{Aut\mkern 2mu}(X)$ of $X$ contains the subgroup $K$ consisting of all bijections $\psi:X\to X$ of $X$ such that  $\sigma\circ\psi=\sigma$ and $\psi|XX=\mathrm{id}$. This subgroup is isomorphic to $\prod_{x\in \sigma(X)}S_{\sigma^{-1}(x)\setminus XX}$.
\end{proposition}

\begin{proof} It is clear that the group $K$ is isomorphic to the product
of permutation groups $\prod_{x\in \sigma(X)}S_{\sigma^{-1}(x)\setminus XX}$.

To see that $K\subset\mathrm{Aut\mkern 2mu}(X)$, it is necessary to check that each bijection $\psi\in K$ is an  automorphism of the semigroup $X$.
Given two elements $x,y\in X$, we shall show that $\psi(xy)=\psi(x)\cdot\psi(y)$. It follows that $xy\in XX$ and hence $\psi(xy)=xy$. Since $\sigma\circ\psi=\sigma$, we obtain that $\sigma(\psi(x))=\sigma(x)$ and $\sigma(\psi(y))=\sigma(y)$. Now Definition~\ref{d:good} ensures that
$$\psi(x)\cdot\psi(y)=x\cdot\psi(y)=xy=\psi(xy),$$
so $\psi\in\mathrm{Aut\mkern 2mu}(X)$.
\end{proof}



A homomorphism $\rho:S\to S$ of a semigroup $S$ is called a {\em homomorphic retraction} if $\rho\circ\rho=\rho$. The functoriality of the superextension in the category of semigroups \cite{G4} ensures that for any homomorphic retraction $\rho:S\to S$ the map $\lambda\rho:\lambda(S)\to\lambda(S)$ is a homomorphic retraction, too.



For a semigroup $X$ by $\Xi_X$ denote the family of all characteristic subsets of $X$.

\begin{theorem}\label{t:Ker} If $\sigma:X\to X$ is an auto-shift of a semigroup $X$, then $\sigma(X)$ is a characteristic ideal of $X$, and for the restriction operator $$R:\mathrm{Aut\mkern 2mu}(X)\to\mathrm{Aut\mkern 2mu}(\sigma(X)),\;\; R:\psi\mapsto\psi|\sigma(X),$$
the kernel of $R$ is equal to $\prod_{x\in\sigma(X)}S_{\sigma^{-1}(x)\setminus\sigma(X)}$ and $R(\mathrm{Aut\mkern 2mu}(X))\subset H\subset G$ where  $$G=\big\{\varphi\in\mathrm{Aut\mkern 2mu}(\sigma(X)):\forall x\in\sigma(X)\;\;\varphi(\sigma^{-1}(x)\cap\sigma(X))=\sigma^{-1}(\varphi(x))\cap\sigma(X)\mbox{ \ and \ } |\sigma^{-1}(x)\setminus\sigma(X)|=|\sigma^{-1}(\varphi(x))\setminus\sigma(X)|\big\}$$
and
$$H:=\big\{\varphi\in G:\forall C\in\Xi_X\;\forall x\in\sigma(X)\;\;|\sigma^{-1}(x)\cap C|=|\sigma^{-1}(\varphi(x))\cap C|\big\}.$$If the auto-shift $\sigma$ is a homomorphic retraction, then $R(\mathrm{Aut\mkern 2mu}(X))=H=G$.
\end{theorem}

\begin{proof} Definition~\ref{d:good} ensures that $XX\subset \sigma(X)$, so $\sigma(X)$ is an ideal in $X$. Since $\sigma$ is an auto-shift, for any automorphism $\psi\in\mathrm{Aut\mkern 2mu}(X)$ we have $\psi\circ\sigma=\sigma\circ\psi$ and hence
$$\psi(\sigma(X))=\psi\circ\sigma(X)=\sigma\circ \psi(X)=
\sigma(X),$$
which means that the ideal $\sigma(X)$ is characteristic in $X$.

Now consider the group $K$ of all bijections $\psi:X\to X$ such that $\sigma\circ\psi=\sigma$ and $\psi|\sigma(X)$ is the identity map of $\sigma(X)$.
It is clear that the group $K$ is isomorphic to the product of permutation groups  $\prod_{x\in \sigma(x)}S_{\sigma^{-1}(x)\setminus\sigma(X)}$.
By Proposition~\ref{p:kernel}, $K\subset\mathrm{Aut\mkern 2mu}(X)$.

It is clear that $K$ is contained in the kernel $\mathrm{Ker}(R)$ of the restriction operator $R:\mathrm{Aut\mkern 2mu}(X)\to\mathrm{Aut\mkern 2mu}(\sigma(X))$. On the other hand, for any automorphism $\psi\in\mathrm{Ker}(R)$ we get $\sigma\circ\psi=\psi\circ\sigma=\sigma$, so $\psi\in K$.

Observe that for any automorphism $\psi$ of $X$ the equality $\psi^{-1}\circ\sigma=\sigma\circ\psi^{-1}$ implies that
$\sigma^{-1}(y)=\sigma^{-1}(\psi(y))$ for any $y\in\sigma(X)$.
Moreover, for any characteristic set $C\in\Xi_X$ we get $\psi(C)=C$ and hence
$$\psi(\sigma^{-1}(x)\cap C)=\psi(\sigma^{-1}(x))\cap \psi(C)=\sigma^{-1}(\psi(x))\cap C$$ and hence  $|\sigma^{-1}(x)\cap C|=|\sigma^{-1}(\psi(x))\cap C|$. Since $\sigma(X)\in\Xi_X$, for the restriction $\varphi:=\psi|\sigma(X)$ of $\psi$ we get $\varphi(\sigma^{-1}(x)\cap\sigma(X))=\sigma^{-1}(x)\cap\sigma(X)$ for every $x\in\sigma(X)$, which means that $\varphi=R(\psi)\in H$ and hence $R(\mathrm{Aut\mkern 2mu}(X))\subset H$.

The inclusion $H\subset G$ follows from $X\setminus \sigma(X)\in\Xi_X$.

Assuming that $\sigma$ is a homomorphic retraction, we shall prove that $R(\mathrm{Aut\mkern 2mu}(X))=H=G$.
It suffices to check that each automorphism $\varphi\in G$ extends to an automorphism of the semigroup $X$.
By the definition of $G$, we can extend the automorphism $\varphi$ of $\sigma(X)$ to a bijection $\bar\varphi$
of $X$ such that $\bar\varphi(\sigma^{-1}(x)\setminus\sigma(X))=\sigma^{-1}(\varphi(x))\setminus\sigma(X)$ for all $x\in\sigma(X)$.
By $\varphi\in G$, we also get $\varphi(\sigma^{-1}(x))\cap\sigma(X)=\sigma^{-1}(\varphi(x))\cap\sigma(X)$. So, $\bar\varphi(\sigma^{-1}(x))=\sigma^{-1}(\varphi(x))$ for any $x\in\sigma(X)$.
This equality implies the equality $\sigma\circ\bar\varphi=\varphi\circ\sigma$. Indeed, for any $z\in X$ and $x=\sigma(z)$ we get $\bar\varphi(z)\in\bar\varphi(\sigma^{-1}(x))=\sigma^{-1}(\varphi(x))$. Applying to this equality the map $\sigma$, we obtain $\sigma\circ\bar\varphi(z)\in\{\varphi(x)\}$ and hence $\sigma\circ\bar\varphi(z)=\varphi(x)=\varphi\circ\sigma(z)$.

Let us show that the bijection $\bar\varphi$ is an automorphism of $X$. Given any elements $x,y\in X$, it suffices to check that $\bar\varphi(xy)=\bar\varphi(x)\cdot\bar\varphi(y)$.
Taking into account that $xy\in XX\subset\sigma(X)$ and $\sigma:X\to\sigma(X)$ is a homomorphic retraction, we conclude that $xy=\sigma(xy)=\sigma(x)\cdot\sigma(y)$ and thus $$
\bar\varphi(xy)=\varphi(xy)=\varphi(\sigma(xy))=\varphi(\sigma(x)\cdot\sigma(y))=
\varphi(\sigma(x))\cdot\varphi(\sigma(y))=\sigma(\bar\varphi(x))\cdot
\sigma(\bar\varphi(y))=\sigma(\bar\varphi(x)\cdot\bar\varphi(y))=
\bar\varphi(x)\cdot
\bar\varphi(y).
$$
\end{proof}


\section{Automorphisms and characteristic ideals of superextensions of finite monogenic semigroups}\label{monsem}

A semigroup $S$ is called {\em monogenic} if it is generated by some element $a\in S$ in the sense that $S=\{a^n\}_{n\in\mathbb N}$. If a monogenic semigroup is
infinite, then it is isomorphic to the additive semigroup $\mathbb N$ of positive integer numbers. A
finite monogenic semigroup $S=\langle a\rangle$ also has simple
structure, see  \cite{Howie}. There are positive integer numbers $r$
and $m$ called the {\em index} and the {\em period} of $S$ such
that
\begin{itemize}
\item $S=\{a,a^2,\dots,a^{r+m-1}\}$ and $r+m-1=|S|$;
\item $a^{r+m}=a^{r}$;
\item $C_m:=\{a^r,a^{r+1},\dots,a^{r+m-1}\}$ is
 a cyclic and maximal subgroup of $S$ with the
neutral element $e=a^n\in C_m$ and generator $a^{n+1}$, where $n\in (m\cdot\mathbb N)\cap\{r,\dots,r+m-1\}$.
\end{itemize}



From now on, we denote by $\mathrm{M\mkern 1mu}_{r,m}$ a finite monogenic semigroup of
index $r$ and period $m$. Consider the shift
$$\sigma:\mathrm{M\mkern 1mu}_{r,m}\to a\mathrm{M\mkern 1mu}_{r,m},\;\;\sigma:x\mapsto ax,$$and observe that for
every $k\in\mathbb N$ the shift $\sigma^k:\mathrm{M\mkern 1mu}_{r,m}\to a^k\mathrm{M\mkern 1mu}_{r,m}$, $\sigma^k:x\mapsto a^kx$, coincides with the $k$th iteration of $\sigma$.

Observe also that the subset $a^k\mathrm{M\mkern 1mu}_{r,m}=\sigma^k(\mathrm{M\mkern 1mu}_{r,m})$ coincides with
the characteristic ideal $\mathrm{M\mkern 1mu}_{r,m}^{\cdot (k+1)}$ of $\mathrm{M\mkern 1mu}_{r,m}$. For $k\ge r$
the characteristic ideal $\mathrm{M\mkern 1mu}_{r,m}^{\cdot k}=a^{k-1}\mathrm{M\mkern 1mu}_{r,m}$ coincides with the minimal
ideal of $\mathrm{M\mkern 1mu}_{r,m}$ and with a maximal subgroup $C_m$ of $\mathrm{M\mkern 1mu}_{r,m}$.

For the idempotent $e=a^n$ of the group
$C_m$ the shift $\sigma^n:\mathrm{M\mkern 1mu}_{r,m}\to a^n\mathrm{M\mkern 1mu}_{r,m}=C_m$ is a homomorphic retraction of $\mathrm{M\mkern 1mu}_{r,m}$ onto
its maximal subgroup $C_m$. This retraction will be denoted by $\rho$.


The following lemma distinguishes some characteristic ideals in the semigroup $\lambda(\mathrm{M\mkern 1mu}_{r,m})$.

\begin{lemma}\label{l:char-i} For every $k\ge 2$ the subsemigroup $\lambda(\mathrm{M\mkern 1mu}^{\cdot k}_{r,m})$ of $\lambda(\mathrm{M\mkern 1mu}_{r,m})$
coincides with the characteristic ideal  $\lambda(\mathrm{M\mkern 1mu}_{r,m})^{\cdot k}$ of the semigroup $\lambda(\mathrm{M\mkern 1mu}_{r,m})$.
\end{lemma}

\begin{proof} By the functoriality of the superextension, the surjective map
$$\sigma^{k-1}:\mathrm{M\mkern 1mu}_{r,m}\to a^{k-1}\mathrm{M\mkern 1mu}_{r,m}=\mathrm{M\mkern 1mu}_{r,m}^{\cdot k},\;\;\sigma^{k-1}:x\mapsto a^{k-1}x,$$
induces the surjective map $\lambda \sigma^{k-1}:\lambda(\mathrm{M\mkern 1mu}_{r,m})\to\lambda(\mathrm{M\mkern 1mu}_{r,m}^{\cdot k})$, $\lambda\sigma^{k-1}:\mathcal A\mapsto a^{k-1}*\mathcal A$.
Then for every maximal linked upfamily $\mathcal B\in\lambda(a^{k-1}\mathrm{M\mkern 1mu}_{r,m})$ we can find a maximal linked upfamily $\mathcal A\in\lambda(\mathrm{M\mkern 1mu}_{k,m})$ with $\mathcal B=a^{k-1}*\mathcal A$ and conclude that $
\mathcal B=a^{k-1}*\mathcal A\in\lambda(\mathrm{M\mkern 1mu}_{k,m})^{\cdot k}$.

On the other hand, the definition of the semigroup operation on $\lambda(\mathrm{M\mkern 1mu}_{r,m})$
implies that $\lambda(\mathrm{M\mkern 1mu}_{r,m})^{\cdot k}\subset \lambda(\mathrm{M\mkern 1mu}_{r,m}^{\cdot k})= \lambda(a^{k-1}\mathrm{M\mkern 1mu}_{r,m})$.
So, $\lambda(\mathrm{M\mkern 1mu}_{r,m}^{\cdot k})=\lambda(\mathrm{M\mkern 1mu}_{r,m})^{\cdot k}$ is a characteristic ideal in $\lambda(\mathrm{M\mkern 1mu}_{r,m})$.
\end{proof}



\begin{theorem}\label{t3.2} Two finite monogenic semigroups are isomorphic if and only if their superextensions are isomorphic.
\end{theorem}

\begin{proof} If two semigroups are isomorphic, then their superextensions are isomorphic by the functoriality of the superextension in the category of semigroups \cite{G4}. Now assume that two monogenic semigroups $\mathrm{M\mkern 1mu}_{i,m}$ and $\mathrm{M\mkern 1mu}_{j,n}$ have isomorphic superextensions. Let $\psi:\lambda(\mathrm{M\mkern 1mu}_{i,m})\to\lambda(\mathrm{M\mkern 1mu}_{j,n})$ be an isomorphism.

Let $C_m$ be the maximal subgroup of $\mathrm{M\mkern 1mu}_{i,m}$. By Lemma~\ref{l:char-i}, the superextension $\lambda(C_m)$ is a characteristic subsemigroup of $\lambda(\mathrm{M\mkern 1mu}_{i,m})$, equal to the intersection $\bigcap_{k\in\mathbb N}\lambda(\mathrm{M\mkern 1mu}_{i,m})^{\cdot k}$ of the characteristic ideals $\lambda(\mathrm{M\mkern 1mu}_{i,m})^{\cdot k}$. Then
$$\psi(\lambda(C_m))=\psi\big({\textstyle \bigcap}_{k\in\mathbb N}\lambda(\mathrm{M\mkern 1mu}_{i,m})^{\cdot k}\big)=\bigcap_{k\in\mathbb N}\psi\big(\lambda(\mathrm{M\mkern 1mu}_{i,m})^{\cdot k}\big)=
\bigcap_{k\in\mathbb N}\big(\psi(\lambda(\mathrm{M\mkern 1mu}_{i,m}))\big)^{\cdot k}=
\bigcap_{k\in\mathbb N}\lambda(\mathrm{M\mkern 1mu}_{j,n})^{\cdot k}=
\lambda(C_n)$$and hence
$|\lambda(C_m)|=|\lambda(C_n)|$.

Observe that two finite sets $X,Y$ have the same cardinality if $|\lambda(X)|=|\lambda(Y)|$.
Indeed, assuming that $|X|<|Y|$ we can choose an injective map $f:X\to Y$ with $f(X)\ne Y$ and obtain an injective
map $\lambda f:\lambda(X)\to\lambda(Y)$ with $\lambda f(\lambda(X))\ne\lambda(Y)$, which implies that $|\lambda(X)|<|\lambda(Y)|$. Now we see that the equality $|\lambda(C_m)|=|\lambda(C_n)|$ implies $n=m$.

On the other hand, the equality $|\lambda(\mathrm{M\mkern 1mu}_{i,m})|=|\lambda(\mathrm{M\mkern 1mu}_{j,n})|$ implies $i+m-1=|\mathrm{M\mkern 1mu}_{i,m}|=|\mathrm{M\mkern 1mu}_{j,n}|=j+n-1$. Since $m=n$, we obtain that $i=j$ and hence $\mathrm{M\mkern 1mu}_{i,n}=\mathrm{M\mkern 1mu}_{j,m}$.
\end{proof}



\begin{proposition}\label{mono1} If $r\ge 3$, then any automorphism $\psi$ of the semigroup $\lambda(\mathrm{M\mkern 1mu}_{r,m})$ has $\psi(x)=x$ for all $x\in \mathrm{M\mkern 1mu}_{r,m}$.
\end{proposition}

\begin{proof} Let $a$ be the (unique) generator of the monogenic semigroup $\mathrm{M\mkern 1mu}_{r,m}$.
By Lemma~\ref{l:char-i}, $a*\lambda(\mathrm{M\mkern 1mu}_{r,m})=\lambda(\mathrm{M\mkern 1mu}_{r,m})^{\cdot 2}$ is a characteristic
ideal in $\lambda(\mathrm{M\mkern 1mu}_{r,m})$. Then for any automorphism $\psi\in\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{r,m}))$ we get
 $$\psi(a)*\lambda(\mathrm{M\mkern 1mu}_{r,m})=\psi(a)*\psi(\lambda(\mathrm{M\mkern 1mu}_{r,m}))=\psi(a*\lambda(\mathrm{M\mkern 1mu}_{r,m}))=a*\lambda(\mathrm{M\mkern 1mu}_{r,m})\ni a^2$$ and
 hence $a^2=\psi(a)*\mathcal A$ for some $\mathcal A\in\lambda(\mathrm{M\mkern 1mu}_{r,m})$. Taking into account that for $r\ge 3$, the equality $a^2=x*y$ has a unique solution $x=y=a$ in $\lambda(\mathrm{M\mkern 1mu}_{r,m})$, we conclude that $\psi(a)=a$ and hence $\psi(x)=x$ for all $x\in\mathrm{M\mkern 1mu}_{r,m}$.
\end{proof}

In the following proposition by $e$ we denote the unique idempotent of the
group $C_m\subset\mathrm{M\mkern 1mu}_{r,m}$ and by $\rho:\mathrm{M\mkern 1mu}_{r,m}\to C_m$, $\rho:x\mapsto ex$, the homomorphic
retraction of $\mathrm{M\mkern 1mu}_{r,m}$ onto $C_m$. The homomorphic retraction $\rho$ induces a homomorphic
retraction $\bar\rho:\lambda(\mathrm{M\mkern 1mu}_{r,m})\to\lambda(C_m)$, $\bar\rho:\mathcal A\mapsto e*\mathcal A$.

\begin{theorem}\label{t:r=2} For $r=2$ the homomorphic retraction $\bar\rho:\lambda(\mathrm{M\mkern 1mu}_{r,m})\to\lambda(C_m)\subset\lambda(\mathrm{M\mkern 1mu}_{r,m})$ has the following properties:
\begin{enumerate}
\item $\mathcal A*\mathcal B=\bar\rho(\mathcal A)*\mathcal B=\mathcal A*\bar\rho(\mathcal B)=\bar\rho(\mathcal A)*\bar\rho(\mathcal B)$ for any $\mathcal A,\mathcal B\in\lambda(\mathrm{M\mkern 1mu}_{r,m})$;
\item $\psi(x)=x$ for any $x\in C_m$ and any $\psi\in\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{r,m}))$;
\item the homomorphic retraction $\bar\rho$ is an auto-good shift of $\lambda(\mathrm{M\mkern 1mu}_{r,m})$;
\item the operator $R:\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{r,m}))\to\mathrm{Aut\mkern 2mu}(\lambda(C_m))$ has kernel isomorphic to $\prod_{\mathcal L\in\lambda(C_m)}S_{\bar\rho^{-1}(\mathcal L)\setminus\{\mathcal L\}}$ and the range $R(\mathrm{Aut\mkern 2mu}(\mathrm{M\mkern 1mu}_{r,m}))=\{\varphi\in\mathrm{Aut\mkern 2mu}(\lambda(C_m)):\forall \mathcal L\in\lambda(C_m)\;\;|\bar\rho^{-1}(\varphi(\mathcal L))|=|\bar\rho^{-1}(\mathcal L)|\}$.
\end{enumerate}
\end{theorem}

\begin{proof}
 1. The first statement follows from the definition of the semigroup operation on $\lambda(\mathrm{M\mkern 1mu}_{r,m})$ and the equality $xy=\rho(x)\cdot y=x\cdot\rho(y)=\rho(x\cdot y)$ holding for any elements $x,y\in\mathrm{M\mkern 1mu}_{r,m}=\mathrm{M\mkern 1mu}_{2,m}$.
\smallskip

2. Fix any automorphism $\psi$ of the semigroup $\lambda(\mathrm{M\mkern 1mu}_{r,m})$.
By Lemma~\ref{l:char-i}, $\lambda(C_m)$ is a characteristic ideal of $\lambda(\mathrm{M\mkern 1mu}_{r,m})$,
so the restriction $\psi|\lambda(C_m)$ is an automorphism of the semigroup $\lambda(C_m)$. In \cite{BG9} it was proved that $\psi(C_m)=C_m$.
Taking into account that $e$ is a unique idempotent of $C_m$, we conclude that $\psi(e)=e$.

Since $\psi$ is an automorphism of $\lambda(\mathrm{M\mkern 1mu}_{r,m})$, for every $\mathcal A\in\lambda(\mathrm{M\mkern 1mu}_{r,m})$ we have
$$\psi(\bar\rho(\mathcal A))=\psi(e*\mathcal A)=\psi(e)*\psi(\mathcal A)=e*\psi(\mathcal A)=\bar\rho(\psi(\mathcal A)),$$
which implies that $\psi(\bar\rho^{-1}(\mathcal B))=\bar\rho^{-1}(\psi(\mathcal B))$ and hence
$|\bar\rho^{-1}(\mathcal B)|=|\bar\rho^{-1}(\psi(\mathcal B))|$ for all $\mathcal B\in\lambda(C_m)$. In particular,
for the generator $a$ of the semigroup $\mathrm{M\mkern 1mu}_{r,m}$ we get $|\bar\rho^{-1}(ae)|=|\bar\rho^{-1}(\psi(ae))|$. Now observe that
$\bar\rho^{-1}(ae)\supset\rho^{-1}(ae)=\{ae,a\}$ and $\bar\rho^{-1}(x)=\rho^{-1}(x)=\{x\}$ for any $x\in C_m\setminus\{ae\}$. This implies that $\psi(ae)=ae$ and hence $\psi(x)=x$ for all $x\in C_m$ (as $ae$ is a generator of the group $C_m$).
\smallskip

3. The third statement follows from the preceding two statements.

4. The fourth statement follows from the preceding statement and Theorem~\ref{t:Ker}.
\end{proof}

\begin{lemma}\label{l:auto-shift} If $r\ge 2$, then the map $$\bar\sigma:\lambda(\mathrm{M\mkern 1mu}_{r,m})\to\lambda(\mathrm{M\mkern 1mu}_{r,m}^{\cdot2}),\;\;\bar\sigma:\mathcal A\mapsto a*\mathcal A$$ is an auto-shift of the semigroup $\lambda(\mathrm{M\mkern 1mu}_{r,m})$.
\end{lemma}

\begin{proof}
It is clear that $\bar\sigma(\lambda(\mathrm{M\mkern 1mu}_{r,m}))=a*\lambda(\mathrm{M\mkern 1mu}_{r,m})=\lambda(\mathrm{M\mkern 1mu}_{r,m})^{\cdot 2}$. Next, we show that $\bar\sigma\circ\psi=\psi\circ \bar\sigma$ for every automorphism $\psi$ of $\lambda(\mathrm{M\mkern 1mu}_{r,m})$.

If $r\ge 3$, then $\psi(a)=a$ by Proposition~\ref{mono1} and hence $$\psi\circ\bar\sigma(\mathcal A)=\psi(a*\mathcal A)=\psi(a)*\psi(\mathcal A)=a*\psi(\mathcal A)=\bar\sigma\circ\psi(\mathcal A)$$for any $\mathcal A\in\lambda(\mathrm{M\mkern 1mu}_{r,m})$.

If $r=2$, then $ax=aex$ for any $x\in\mathrm{M\mkern 1mu}_{r,m}$. By Theorem~\ref{t:r=2}, $\psi(ae)=ae$ and then
$$\psi\circ\bar\sigma(\mathcal A)=\psi(a*\mathcal A)=\psi(ae*\mathcal A)=\psi(ae)*\psi(\mathcal A)=ae*\psi(\mathcal A)=a*\psi(\mathcal A)=
\bar\sigma\circ\psi(\mathcal A)$$for any $\mathcal A\in\lambda(\mathrm{M\mkern 1mu}_{r,m})$.

It remains to prove that for any $\mathcal A,\mathcal A'\in\lambda(\mathrm{M\mkern 1mu}_{r,m})$ with $\bar\sigma(\mathcal A)=\bar\sigma(\mathcal A')$ the equalities $\mathcal A*\mathcal B=\mathcal A'*\mathcal B$ and $\mathcal B*\mathcal A=\mathcal B*\mathcal A'$ hold for all $\mathcal B\in\lambda(\mathrm{M\mkern 1mu}_{r,m})$.

It follows from $a*\mathcal A=\bar\sigma(\mathcal A)=\bar\sigma(\mathcal A')=a*\mathcal A'$ that $a^n*\mathcal A=a^n*\mathcal A'$ for all $n\in\mathbb N$ and hence $x*\mathcal A=x*\mathcal A'$
for all $x\in\mathrm{M\mkern 1mu}_{r,m}$. To see that $\mathcal B*\mathcal A=\mathcal B*\mathcal A'$, take any set $C\in\mathcal B*\mathcal A$ and find
a set $B\in\mathcal B$ and a family $(A_b)_{b\in B}\in \mathcal A^B$ such that $\bigcup_{b\in B}bA_b\subset C$.
For every $b\in B$, we can use the equality $bA_b\in b*\mathcal A=b*\mathcal A'$ to find a set $A'_b\in \mathcal A'$ such that $bA_b'\subset bA_b$.
Then $$\mathcal B*\mathcal A'\ni\bigcup_{b\in B}bA_b'\subset\bigcup_{b\in B}bA_b\subset C$$and hence $C\in\mathcal B*\mathcal A'$.
By analogy we can prove that $\mathcal B*\mathcal A'\subset\mathcal B*\mathcal A$.

Next, we prove that $\mathcal A*\mathcal B=\mathcal A'*\mathcal B$. Given any element $C\in\mathcal A*\mathcal B$, find a set $A\in\mathcal A$ and a family $\{B_x\}_{x\in A}\subset\mathcal B$ such that $\bigcup_{x\in A}xB_x\subset C$. Since $aA\in a*\mathcal A=a*\mathcal A'$, there exists a set $A'\in\mathcal A'$ such that $aA'\subset aA$. For any $y\in A'$ find $x\in A$ with $ay=ax$ and put $B_{y}:=B_x$. We claim that $\bigcup_{y\in A'}yB_y\subset C$. Indeed, for any $z\in \bigcup_{y\in A'}yB_y$ we can find $y\in A'$ and $b\in B_y$ such that $z=yb\in yB_y$. For the element $y$ there exists an element $x\in A$ such that $ax=ay$ and $B_y=B_x$. The equality $ax=ay$ implies $a^kx=a^ky$ for all $k\in\mathbb N$. In particular, $bx=by$. Then $z=yb=by=bx=xb\in xB_x\subset C$ and hence $\mathcal A'*\mathcal B\ni\bigcup_{y\in A'}B_y\subset C$, which implies $C\in\mathcal A'*\mathcal B$ and $\mathcal A*\mathcal B\subset \mathcal A'*\mathcal B$. By analogy we can prove that $\mathcal A'*\mathcal B\subset \mathcal A*\mathcal B$ and thus $\mathcal A'*\mathcal B=\mathcal A*\mathcal B$.
\end{proof}

\begin{corollary}\label{c:good} If $r\ge 2$, then for any $k\in\mathbb N$ the map $$\bar\sigma_k:\lambda(\mathrm{M\mkern 1mu}^{\cdot k}_{r,m})\to\lambda(\mathrm{M\mkern 1mu}_{r,m}^{\cdot (k+1)}),\;\;\bar\sigma_k:\mathcal A\mapsto a*\mathcal A,$$ is a good shift of the semigroup $\lambda(\mathrm{M\mkern 1mu}_{r,m}^{\cdot k})$.
\end{corollary}

Lemma~\ref{l:auto-shift} and Theorem~\ref{t:Ker} have the following:

\begin{corollary}\label{c:Ker} Assume that $r\ge 2$. The operator $R:\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{r,m}))\to\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{r,m}^{\cdot2}))$
has kernel isomorphic to $\prod_{\mathcal L\in\lambda(\mathrm{M\mkern 1mu}_{r,m}^{\cdot 2})}S_{\bar\sigma^{-1}(\mathcal L)\setminus\lambda(\mathrm{M\mkern 1mu}_{r,m}^{\cdot2})}$
and range $R(\mathrm{Aut\mkern 2mu}(\mathrm{M\mkern 1mu}_{r,m}))\subset H$ where
$$
\begin{aligned}
H=\big\{\varphi\in\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{r,m}^{\cdot 2})):\ &\forall \mathcal L\in\lambda(\mathrm{M\mkern 1mu}_{r,m}^{\cdot 2})\;\;\varphi(\bar\sigma^{-1}(\mathcal L)\cap\lambda(\mathrm{M\mkern 1mu}_{r,m}^{\cdot2}))=\bar\sigma^{-1}(\mathcal L)\cap\lambda(\mathrm{M\mkern 1mu}_{r,m}^{\cdot2})\mbox{ and }\\
&\forall C\in\Xi_{\lambda(\mathrm{M\mkern 1mu}_{r,m})}\;|\bar\sigma^{-1}(\varphi(\mathcal L))\cap C|=|\bar\sigma^{-1}(\mathcal L)\cap C|\big\}.
\end{aligned}
$$
\end{corollary}



\section{Automorphism groups of superextensions of monogenic semigroups of cardinality $\le 5$}\label{s4}

In this section we shall describe the structure of the automorphism groups of superextensions of all monogenic semigroups $\mathrm{M\mkern 1mu}_{r,m}$ of cardinality $|\mathrm{M\mkern 1mu}_{r,m}|\le 5$.


\subsection{The semigroups $\lambda(\mathrm{M\mkern 1mu}_{1,1})$, $\lambda(\mathrm{M\mkern 1mu}_{1,2})$ and $\lambda(\mathrm{M\mkern 1mu}_{2,1})$} For any $r,m\in\mathbb N$ with $r+m\le 3$, the monogenic semigroup $\mathrm{M\mkern 1mu}_{r,m}$ has cardinality $|\mathrm{M\mkern 1mu}_{r,m}|\le 2$. Consequently, $\lambda(\mathrm{M\mkern 1mu}_{r,m})=\mathrm{M\mkern 1mu}_{r,m}$ and $\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{r,m}))=\mathrm{Aut\mkern 2mu}(\mathrm{M\mkern 1mu}_{r,m})\cong C_1$.

\subsection{The semigroups $\lambda(\mathrm{M\mkern 1mu}_{1,3})$, $\lambda(\mathrm{M\mkern 1mu}_{2,2})$ and $\lambda(\mathrm{M\mkern 1mu}_{3,1})$}
For  a monogenic semigroup $\mathrm{M\mkern 1mu}_{r,m}$ with $m+r=4$ and generator $a$, the superextension $\lambda(\mathrm{M\mkern 1mu}_{r,m})$  consists of three principal ultrafilters $a,a^2,a^3$
 and the maximal linked family
$\triangle=\langle\{a,a^2\},\{a,a^3\},\{a^2,a^3\}\rangle$.

Taking into account that $\mathrm{M\mkern 1mu}_{1,3}$ is isomorphic to $C_3$ and $\mathrm{Aut\mkern 2mu}(\lambda(C_3))\cong C_2$ (see Table~\ref{tab:dedekind_gr}), we conclude that $\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{1,3}))\cong\mathrm{Aut\mkern 2mu}(\lambda(C_3))\cong C_2$.

By Proposition~\ref{mono1}, for any $i\in\{1,2,3\}$  and $\psi\in\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{3,1}))$
we have $\psi(a^i)=a^i$. Then  $\psi(\triangle)=\triangle$ and hence $\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{3,1}))\cong C_1$.

To recognize the automorphism group of $\lambda(\mathrm{M\mkern 1mu}_{2,2})$, consider the homomorphic
retraction $$\bar\rho:\lambda(\mathrm{M\mkern 1mu}_{2,2})\to\lambda(\mathrm{M\mkern 1mu}_{2,2}^{\cdot2})=\{a^2,a^3\},\;\;\bar\rho:\mathcal A\mapsto a^2\mathcal A$$and
observe that it has fibers: $\bar\rho^{-1}(a^2)=\{a^2\}$ and $\bar\rho^{-1}(a^3)=\{a,a^3,\triangle\}$.
Since the automorphism group $\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{2,2}^{\cdot 2}))\cong\mathrm{Aut\mkern 2mu}(\lambda (C_2))$ is trivial, we can apply Theorem~\ref{t:r=2}
and conclude that the automorphism group $\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{2,2}))$ is isomorphic to $S_{\bar\rho^{-1}(a^3)\setminus\{a^3\}}=S_{\{a,\triangle\}}\cong C_2$.


\subsection{The semigroup $\lambda(\mathrm{M\mkern 1mu}_{1,4})$}
The semigroup $M_{1,4}$ is isomorphic to the cyclic group $C_4$. Therefore, $$\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{1,4}))\cong\mathrm{Aut\mkern 2mu}(\lambda(C_4))\cong C_2\times C_2,$$
according to \cite{BG9} (see also Table~\ref{tab:dedekind_gr}).

\subsection{The semigroup $\lambda(\mathrm{M\mkern 1mu}_{2,3})$}\label{ss:M23}
Consider the semigroup $\mathrm{M\mkern 1mu}_{2,3}=\{a,a^2,a^3,a^4\ |\ a^5=a^2\}$ generated by the element $a$. Its superextension $\lambda(\mathrm{M\mkern 1mu}_{2,3})$ contains $12$ elements of the form:
$$a^i,\;\;\triangle_i=\langle A:A\subset \mathrm{M\mkern 1mu}_{2,3}\setminus\{a^i\},\;|A|=2\rangle,\mbox{ and }\;\;
\square_i=\langle\mathrm{M\mkern 1mu}_{2,3}\setminus\{a^i\},\;\{a^i,x\}:x\in\mathrm{M\mkern 1mu}_{2,3}\setminus\{a^i\}\rangle$$
where $i\in\{1,2,3,4\}$.


Let $e=a^3$ be the idempotent of the subgroup $C_3$ and $\rho:\mathrm{M\mkern 1mu}_{2,3}\to C_3$, $\rho:x\mapsto ex$,
be the homomorphic retraction of $\mathrm{M\mkern 1mu}_{2,3}$ onto $C_3$. The map $\rho$ induces a homomorphic
retraction $\bar\rho:\lambda(\mathrm{M\mkern 1mu}_{2,3})\to\lambda(C_3)=\{a^2,a^3,a^4,\triangle_1\}$, defined by $\bar\rho(\mathcal A)=a^3*\mathcal A$ for $\mathcal A\in\lambda(\mathrm{M\mkern 1mu}_{2,3})$.
By routine calculations we can show that the map $\bar\rho$ has fibers:
\begin{itemize}
\item $\bar\rho^{-1}(a^2)=\{a^2\}$;
\item $\bar\rho^{-1}(a^3)=\{a^3\}$;
\item $\bar\rho^{-1}(a^4)=\{a,\;a^4,\triangle_2,\triangle_3,\square_1,\square_4\}$;
\item $\bar\rho^{-1}(\triangle_1)=\{\triangle_1,\;\triangle_4,\square_2,\square_3\}$.
\end{itemize}

Applying Theorem~\ref{t:r=2}, we can show that the automorphism group $\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{2,3}))$ of $\lambda(\mathrm{M\mkern 1mu}_{2,3})$ is isomorphic to
$$S_{(\bar\rho)^{-1}(\triangle_1)\setminus\{\triangle_1\}}\times S_{(\bar\rho)^{-1}(a^4)\setminus\{a^4\}}\cong S_3\times S_5.$$



\subsection{The semigroup $\lambda(\mathrm{M\mkern 1mu}_{3,2})$}

In this section we consider the monogenic semigroup $\mathrm{M\mkern 1mu}_{3,2}=\{a,a^2,a^3,a^4\}$ generated by an element $a$ such that $a^5=a^3$. The semigroup $\mathrm{M\mkern 1mu}_{3,2}$ contains the characteristic ideals $\mathrm{M\mkern 1mu}_{3,2}^{\cdot 2}=\{a^2,a^3,a^4\}$ and $\mathrm{M\mkern 1mu}_{3,2}^{\cdot3}=\{a^3,a^4\}=C_2$.

Using the notations from the preceding subsection, we can see that the superextensions of the semigroups $\mathrm{M\mkern 1mu}_{3,2}^{\cdot k}$ contain the following elements:
\begin{itemize}
\item $\lambda(\mathrm{M\mkern 1mu}_{3,2})=\{a^i,\triangle_i,\square_i:1\le i\le 4\}$;
\item $\lambda(\mathrm{M\mkern 1mu}_{3,2}^{\cdot 2})=\{a^2,a^3,a^4,\triangle_1\}$;
\item $\lambda(\mathrm{M\mkern 1mu}_{3,2}^{\cdot3})=\lambda(C_2)=C_2=\{a^3,a^4\}$.
\end{itemize}

By Lemma~\ref{l:auto-shift}, the map $\bar\sigma:\lambda(\mathrm{M\mkern 1mu}_{3,2})\to\lambda(\mathrm{M\mkern 1mu}_{3,2}^{\cdot2})$, \ $\bar\sigma:\mathcal A\mapsto a*\mathcal A$, is an auto-shift of the semigroup $\lambda(\mathrm{M\mkern 1mu}_{3,2})$.

By  routine calculations it can be shown that the map $\bar\sigma$ has the following  fibers:
\begin{itemize}
\item $\bar\sigma^{-1}(a^2)=\{a\}$;
\item $\bar\sigma^{-1}(a^3)=\{a^2,a^4,\triangle_1,\triangle_3,\square_2,\square_4\}$;
\item $\bar\sigma^{-1}(a^4)=\{a^3\}$;
\item $\bar\sigma^{-1}(\triangle_1)=\{\triangle_2,\triangle_4,\square_1,\square_3\}$.
\end{itemize}

By Corollary~\ref{c:good}, the restriction $\bar\sigma_2=\bar\sigma|\lambda(\mathrm{M\mkern 1mu}_{3,2}^{\cdot 2})$ is a good shift of the
semigroup $\lambda(\mathrm{M\mkern 1mu}_{3,2}^{\cdot 2})$. This shift has a unique fiber $\bar\sigma_2^{-1}(a^3)=\{a^4,a^2,\triangle_1\}$
containing more than one point. By Proposition~\ref{p:kernel}, the automorphism group $\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{3,2}^{\cdot2}))$ contains
the permutation group $S_{\{a^2,\triangle_1\}}\cong C_2$. Taking into account that $C_2$ is a characteristic subgroup
of $\lambda(\mathrm{M\mkern 1mu}_{3,2}^{\cdot2})$ with trivial automorphism group, we conclude that $\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{3,2}^{\cdot2}))=S_{\{a^2,\triangle_1\}}\cong C_2$.

By Corollary~\ref{c:Ker}, the restriction operator $R:\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{3,2}))\to\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{3,2}^{\cdot2}))\cong C_2$ has kernel isomorphic to
$$S_{\bar\sigma^{-1}(a^3)\setminus\lambda(\mathrm{M\mkern 1mu}_{3,2}^{\cdot2})}\times S_{\bar\sigma^{-1}(\triangle_1)\setminus\lambda(\mathrm{M\mkern 1mu}_{3,2}^{\cdot2})}\cong S_3\times S_4.$$
We claim that the subgroup $R(\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{3,2})))$ of $\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{3,2}^{\cdot2}))=S_{\{a^2,\triangle_1\}}$ is trivial.
Indeed, each automorphism $\varphi\in R(\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{3,2})))$ is trivial on the subgroup
$C_2=\{a^3,a^4\}$, so $\varphi(a^2)\in\{a^2,\triangle_1\}$. By Corollary~\ref{c:Ker},
$$1=|\bar\sigma^{-1}(a^2)|=|\bar\sigma^{-1}(\varphi(a^2))|<4=|\bar\sigma^{-1}
(\triangle_1)|,$$ which implies that $\varphi(a^2)=a^2$ and $\varphi$ is the identity automorphism of $\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{3,2}^{\cdot2}))$.
Consequently, the range of the restriction operator $R$ is trivial and the group $\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{3,2}))$
coincides with the kernel of $R$ and hence is isomorphic to $S_3\times S_4$.


\subsection{The semigroup $\lambda(\mathrm{M\mkern 1mu}_{4,1})$}

In this subsection we consider the monogenic semigroup $\mathrm{M\mkern 1mu}_{4,1}=\{a,a^2,a^3,a^4\}$ generated by an element $a$ such that $a^5=a^4$.
The semigroup $\mathrm{M\mkern 1mu}_{4,1}$ contains the characteristic ideals $\mathrm{M\mkern 1mu}_{4,1}^{\cdot 2}=\{a^2,a^3,a^4\}$,  $\mathrm{M\mkern 1mu}_{4,1}^{\cdot3}=\{a^3,a^4\}$, and $\mathrm{M\mkern 1mu}_{4,1}^{\cdot 4}=\{a^4\}$.

Using the notations from the preceding subsection, we can see that the superextensions of the semigroups $\mathrm{M\mkern 1mu}_{4,1}^{\cdot k}$ contain the following elements:
\begin{itemize}
\item $\lambda(\mathrm{M\mkern 1mu}_{4,1})=\{a^i,\triangle_i,\square_i:1\le i\le 4\}$;
\item $\lambda(\mathrm{M\mkern 1mu}_{4,1}^{\cdot 2})=\{a^2,a^3,a^4,\triangle_1\}$;
\item $\lambda(\mathrm{M\mkern 1mu}_{4,1}^{\cdot3})=\mathrm{M\mkern 1mu}_{4,1}^{\cdot 3}=\{a^3,a^4\}$;
\item $\lambda(\mathrm{M\mkern 1mu}_{4,1}^{\cdot4})=\mathrm{M\mkern 1mu}_{4,1}^{\cdot4}=\{a^4\}$.
\end{itemize}
It is clear that the semigroups $\lambda(\mathrm{M\mkern 1mu}_{4,1}^{\cdot4})$ and $\lambda(\mathrm{M\mkern 1mu}_{4,1}^{\cdot3})$ have trivial automorphism groups.

Taking into account that $xy=a^4$ for any $x,y\in\lambda(\mathrm{M\mkern 1mu}_{4,1}^{\cdot2})=\{a^2,a^3,a^4,\triangle_1\}$, we conclude that the automorphism group $\mathrm{Aut\mkern 2mu}(\mathrm{M\mkern 1mu}_{4,1}^{\cdot2})$ of the semigroup $\mathrm{M\mkern 1mu}_{4,1}^{\cdot 2}$ is isomorphic to the symmetric group $S_3$.

By Lemma~\ref{l:auto-shift}, the map $$\bar\sigma:\lambda(\mathrm{M\mkern 1mu}_{4,1})\to\lambda(\mathrm{M\mkern 1mu}_{4,1}^{\cdot2}),\;\;\bar\sigma:\mathcal A\mapsto a*\mathcal A$$
is an auto-shift of the semigroup $\lambda(\mathrm{M\mkern 1mu}_{4,1})$.
By  routine calculations it can be shown that the map $\bar\sigma$ has the following fibers:
\begin{itemize}
\item $\bar\sigma^{-1}(a^2)=\{a\}$;
\item $\bar\sigma^{-1}(a^3)=\{a^2\}$;
\item $\bar\sigma^{-1}(a^4)=\{a^3,a^4,\triangle_1,\triangle_2,\square_3,\square_4\}$;
\item $\bar\sigma^{-1}(\triangle_1)=\{\triangle_3,\triangle_4,\square_1,\square_2\}$.
\end{itemize}

By Corollary~\ref{c:Ker}, the restriction operator $R:\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{4,1}))\to\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{4,1}^{\cdot2}))$ has kernel isomorphic to
$$S_{\bar\sigma^{-1}(a^4)\setminus\lambda(\mathrm{M\mkern 1mu}_{4,1}^{\cdot2})}\times S_{\bar\sigma^{-1}(\triangle_1)\setminus\lambda(\mathrm{M\mkern 1mu}_{4,1}^{\cdot2})}=S_{\{\triangle_2,\square_3,\square_4\}}\times S_{\{\triangle_3,\triangle_4,\square_1,\square_2\}}\cong S_3\times S_4.$$
By Proposition~\ref{mono1}, for any automorphism $\psi$ of $\lambda(\mathrm{M\mkern 1mu}_{4,1})$,
we get $\psi(a^k)=a^k$ for any $k\in\mathbb N$. Taking into account that $\lambda(\mathrm{M\mkern 1mu}_{4,1}^{\cdot2})=\{a^2,a^3,a^4,\triangle_1\}$ is
a characteristic ideal in $\lambda(\mathrm{M\mkern 1mu}_{4,1})$, we conclude that $\psi(\triangle_1)=\triangle_1$.
Consequently, the range of the restriction operator $R$ is trivial and the group $\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{4,1}))$
coincides with the kernel of $R$ and hence is isomorphic to $S_3\times S_4$.

\subsection{The semigroup $\lambda(\mathrm{M\mkern 1mu}_{1,5})$} The monogenic semigroup $\mathrm{M\mkern 1mu}_{1,5}$ is isomorphic to the cyclic group $C_5$.
It was proved in~\cite{BG9} that the automorphism group $\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{1,5}))\cong\mathrm{Aut\mkern 2mu}(\lambda(C_5))$ is isomorphic to $C_4$.

\subsection{The semigroup $\lambda(\mathrm{M\mkern 1mu}_{2,4})$}\label{ss:M24} To describe the structure of the automorphism group $\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{2,4}))$, we shall apply Theorem~\ref{t:r=2}.

Consider the homomorphic retraction
$$\bar\rho:\lambda(\mathrm{M\mkern 1mu}_{2,4})\to\lambda(\mathrm{M\mkern 1mu}_{2,4}^{\cdot 2}),\;\bar\rho:\mathcal L\mapsto e*\mathcal L,$$ where $e=a^4\in C_4\subset\mathrm{M\mkern 1mu}_{2,4}$ is the unique idempotent of the semigroup $\mathrm{M\mkern 1mu}_{2,4}$.

To describe the fibers of the retraction $\bar\rho$ we first introduce a convenient notation for all
81 elements of the superextension $\lambda(\mathrm{M\mkern 1mu}_{2,4})$. We recall that $\mathrm{M\mkern 1mu}_{2,4}=\{a,a^2,a^3,a^4,a^5\}$ and $C_4=\{a^2,a^3,a^4,a^5\}$ where $a^6=a^2$.

The 81 elements of the semigroup $\lambda(\mathrm{M\mkern 1mu}_{2,4})$ will be denoted as follows:
\begin{itemize}
\item $a^i$ for $i\in\{1,2,3,4,5\}$;
\item $\bigcirc:=\{A\subset \mathrm{M\mkern 1mu}_{2,4}:|A|\ge3\}$;
\item $\Theta_{ij}:=\langle\{a^i,a^j\},A:A\subset \mathrm{M\mkern 1mu}_{2,4},\;|A|=3,\;|A\cap\{a^i,a^j\}|=1\}\rangle$ for $1\le i<j\le 5$;
\item $\triangle_{ijk}:=\langle \{a^i,a^j\},\{a^i,a^k\},\{a^j,a^k\}\rangle$ for numbers $1\le i<j<k\le 5$;
\item $\Lambda^i:=\langle \mathrm{M\mkern 1mu}_{2,4}\setminus\{a^i\}, \{a^i,x\}:x\in \mathrm{M\mkern 1mu}_{2,4}\setminus\{a^i\}\rangle$ for $i\in\{1,2,3,4,5\}$;
\item $\diamondsuit^{n}_{ijk}:=\langle\{a^n,a^i\},\{a^n,a^j\},\{a^n,a^k\},\{a^i,a^j,a^k\}
    \rangle$ for numbers $1\le i<j<k\le 5$ and $n\in\{1,2,3,4,5\}\setminus \{i,j,k\}$;
\item $\Lambda^i_{jk}:=\big\langle\{a^i,a^j\},\{a^i,a^k\},A:A\subset \mathrm{M\mkern 1mu}_{2,4},\;|A|=3,\;A\cap\{a^i,a^j,a^k\}\in\big\{\{a^i\},\{a^j,a^k\}\big\}\big\rangle$ for $1\le j<k\le 5$ and $i\in \{1,2,3,4,5\}\setminus \{j,k\}$.
\end{itemize}
Observe that for the subgroup $C_4=\{a^2,a^3,a^4,a^5\}\subset\mathrm{M\mkern 1mu}_{2,4}$ we have
$$\lambda(C_4)=C_4\cup\{\triangle_{234},\triangle_{235},\triangle_{245},\triangle_{345},\diamondsuit^{2}_{345},\diamondsuit^3_{245},\diamondsuit^4_{235},\diamondsuit^5_{234}\}.$$
By routine calculations it can be shown that the map $\bar\rho:\lambda(\mathrm{M\mkern 1mu}_{2,4})\to\lambda(C_4)$ has the following fibers:
\begin{itemize}
\item $(\bar\rho)^{-1}(x)=\{x\}$ for $x\in\{a^2,a^3,a^4,\triangle_{234}\}$;
\item $(\bar\rho)^{-1}(a^5)=\{a^5,\;a,\triangle_{125},\triangle_{135},\triangle_{145},\diamondsuit^1_{235},
\diamondsuit^1_{245},\diamondsuit^1_{345},\diamondsuit^5_{123},\diamondsuit^5_{124}, \diamondsuit^5_{134},\Lambda^1,\Lambda^5,\Theta_{15},\Lambda^1_{25}, \Lambda^1_{35}, \Lambda^1_{45}, \Lambda^5_{12},\Lambda^5_{13},\Lambda^5_{14}\}$;
\item $(\bar\rho)^{-1}(\triangle_{235})=\{\triangle_{235},\;\triangle_{123},
    \diamondsuit^2_{135},\diamondsuit^3_{125},
    \Theta_{23},\Lambda^2_{13},\Lambda^2_{35},\Lambda^3_{12}, \Lambda^3_{25}\}$;
\item $(\bar\rho)^{-1}(\triangle_{245})=\{\triangle_{245},\;\triangle_{124},\diamondsuit^2_{145},
\diamondsuit^4_{125},\Theta_{24},\Lambda^2_{14},\Lambda^2_{45}, \Lambda^4_{12},\Lambda^4_{25}\}$;
\item $(\bar\rho)^{-1}(\triangle_{345})=\{\triangle_{345},\;\triangle_{134},\diamondsuit^3_{145},\diamondsuit^4_{135},\Theta_{34},\Lambda^3_{14},\Lambda^3_{45},\Lambda^4_{13}, \Lambda^4_{35}\}$;
\item $(\bar\rho)^{-1}(\diamondsuit^2_{345})=\{\diamondsuit^2_{345},\;\diamondsuit^2_{134}, \Lambda^2, \Lambda^2_{34}\}$;
\item $(\bar\rho)^{-1}(\diamondsuit^3_{245})=\{\diamondsuit^3_{245},\;\diamondsuit^3_{124},\Lambda^3, \Lambda^3_{24}\}$;
\item $(\bar\rho)^{-1}(\diamondsuit^4_{235})=\{\diamondsuit^4_{235},\;\diamondsuit^4_{123},\Lambda^4,
    \Lambda^4_{23}\}$;
\item $(\bar\rho)^{-1}(\diamondsuit^5_{234})=\{\diamondsuit^5_{234},\;\diamondsuit^1_{234}, \Lambda^1_{23},\Lambda^1_{24},\Lambda^1_{34},\Lambda^2_{15}, \Lambda^3_{15},\Lambda^4_{15},\Lambda^5_{23},\Lambda^5_{24},\Lambda^5_{34},
\Theta_{12},\Theta_{13},\Theta_{14},\Theta_{25},\Theta_{35},
\Theta_{45},\bigcirc\}$.
\end{itemize}


By Theorem~\ref{t:r=2}, for any automorphism $\psi$ of $\lambda(M_{2,4})$ the restriction $\psi|\lambda(C_4)$ is
an automorphism of $\lambda(C_4)$ such that $\psi(x)=x$ for all $x\in C_4$. The description of
the automorphism group of $\lambda(C_4)$ given in~\cite{BG9} ensures that $\psi(\diamondsuit^{n}_{ijk})=\diamondsuit^n_{ijk}$
for any $\diamondsuit^n_{ijk}\in\lambda(C_4)$. Taking into account that
$|(\bar\rho)^{-1}(\triangle_{234})|=1<|(\bar\rho)^{-1}(\triangle_{ijk})|$ for
any $\triangle_{ijk}\in\lambda(C_4)\setminus\{\triangle_{234}\}$, we conclude that $\psi(\triangle_{234})=\psi(\triangle_{234})$
and hence $\psi(x\triangle_{234})=x\triangle_{234}$ for any $x\in C_4$. Since $\{x\triangle_{234}:x\in C_4\}=\{\triangle_{ijk}:2\le i<j<k\le 5\}$,
we conclude that $\psi|\lambda(C_4)$ is the identity automorphism of the semigroup $\lambda(C_4)$.

By Theorem~\ref{t:r=2}, the automorphism group $\mathrm{Aut\mkern 2mu}(\lambda(M_{2,4}))$ is isomorphic
to $$\prod_{\mathcal L\in\lambda(C_4)}S_{(\bar\rho)^{-1}(\mathcal L)\setminus\{\mathcal L\}}\cong S_3^3\times S_8^3\times S_{17}\times S_{19}.$$


\subsection{The semigroup $\lambda(\mathrm{M\mkern 1mu}_{3,3})$}

In this section we recognize the automorphism group of the superextension of the semigroup $\mathrm{M\mkern 1mu}_{3,3}=\{a,a^2,a^3,a^4,a^5\}$.
This semigroup is generated by an element $a$ such that $a^6=a^3$. Observe that the ideal $\mathrm{M\mkern 1mu}_{3,3}^{\cdot2}=\{a^2,a^3,a^4,a^5\}$
is isomorphic to the monogenic semigroup $\mathrm{M\mkern 1mu}_{2,3}=\{b,b^2,b^3,b^4\}$ as for the element $b=a^2$ we get $\{b,b^2,b^3,b^4\}=\{a^2,a^4,a^3,a^5\}$.

As shown in Subsection~\ref{ss:M23}, the automorphism group $\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{2,3}))$ consists of all bijections
$\psi$ of $\lambda(\mathrm{M\mkern 1mu}_{2,3})$ such that
\begin{itemize}
\item $\psi(x)=x$ for any $x\notin\{\triangle_4,\square_2,\square_3\}\cup\{b,\triangle_2,\triangle_3,\square_1,\square_4\}$;
\item the sets $\{\triangle_4,\square_2,\square_3\}$ and $\{b,\triangle_2,\triangle_3,\square_1,\square_4\}$ are $\psi$-invariant.
 \end{itemize}
Taking into account the isomorphism $\{b,b^2,b^3,b^4\}\to\{a^2,a^4,a^3,a^5\}$ between the semigroups $\mathrm{M\mkern 1mu}_{2,3}$ and $\mathrm{M\mkern 1mu}_{3,3}^{\cdot2}$, we obtain the following fact.

\begin{claim}\label{cl:M33} The automorphism group of the semigroup $\lambda(\mathrm{M\mkern 1mu}_{3,3}^{\cdot2})$ consists of bijections $\psi$ of $\lambda(\mathrm{M\mkern 1mu}_{3,3}^{\cdot2})$ such that
\begin{itemize}
\item $\psi(x)=x$ for any $x\not\in\{\triangle_{234},\diamondsuit^4_{235},\diamondsuit^3_{245}\}\cup \{a^2,\triangle_{235},\triangle_{245},\diamondsuit^2_{345},\diamondsuit^5_{234}\}$;
\item the sets $\{\triangle_{234},\diamondsuit^4_{235},\diamondsuit^3_{245}\}$ and
$\{a^2,\triangle_{235},\triangle_{245},\diamondsuit^2_{345},\diamondsuit^5_{234}\}$ are $\psi$-invariant.
\end{itemize}
\end{claim}
By Lemma~\ref{l:auto-shift}, the map $\bar\sigma:\lambda(\mathrm{M\mkern 1mu}_{3,3})\to\lambda(\mathrm{M\mkern 1mu}_{3,3}^{\cdot2})$, $\bar \sigma:\mathcal A\mapsto a*\mathcal A$,
is an auto-shift of the semigroup $\lambda(\mathrm{M\mkern 1mu}_{3,3})$.
By  routine calculations, it can be shown that the map $\bar\sigma$ has the following fibers:
\begin{itemize}
\item $\bar\sigma^{-1}(a^2)=\{a\}$, $\bar\sigma^{-1}(a^4)=\{a^3\}$, $\bar\sigma^{-1}(a^5)=\{a^4\}$;
\item $\bar\sigma^{-1}(a^3)=\{a^2,a^5,\triangle_{125},\triangle_{235},\triangle_{245},\Lambda^2_{15},\Lambda^2_{35},
\Lambda^2_{45},\Lambda^5_{12},\Lambda^5_{23},\Lambda^5_{24},\diamondsuit^2_{135}, \diamondsuit^2_{145}, \diamondsuit^2_{345},
\diamondsuit^5_{123},\diamondsuit^5_{124}, \diamondsuit^5_{234},\Theta_{25},\Lambda^2,\Lambda^5\}$;
\item $\bar\sigma^{-1}(\triangle_{245})=\{\triangle_{134}\}$;
\item $\bar\sigma^{-1}(\triangle_{234})=\{\triangle_{123},\triangle_{135},\Lambda^1_{23},\Lambda^1_{35},\Lambda^3_{12},\Lambda^3_{15},
\diamondsuit^1_{235},\diamondsuit^3_{125}, \Theta_{13}\}$;
\item $\bar\sigma^{-1}(\triangle_{235})=\{\triangle_{124},\triangle_{145}, \Lambda^1_{24}, \Lambda^1_{45}, \Lambda^4_{12},\Lambda^4_{15},  \diamondsuit^1_{245},
\diamondsuit^4_{125}, \Theta_{14}\}$;
\item $\bar\sigma^{-1}(\triangle_{345})=\{\triangle_{234},\triangle_{345},\Lambda^3_{24},\Lambda^3_{45},\Lambda^4_{23},
\Lambda^4_{35},\diamondsuit^3_{245},\diamondsuit^4_{235},\Theta_{34}\}$;
\item $\bar\sigma^{-1}(\diamondsuit^2_{345})= \{\Lambda^1_{34}, \diamondsuit^1_{234}, \diamondsuit^1_{345},\Lambda^1\}$;
\item $\bar\sigma^{-1}(\diamondsuit^4_{235})=\{\Lambda^3_{14},\diamondsuit^3_{124}, \diamondsuit^3_{145},\Lambda^3\}$;
\item $\bar\sigma^{-1}(\diamondsuit^5_{234})=\{\Lambda^4_{13},\diamondsuit^4_{123}, \diamondsuit^4_{135},\Lambda^4\}$;
\item $\bar\sigma^{-1}(\diamondsuit^3_{245})=\{\Lambda^1_{25},\Lambda^2_{13},\Lambda^2_{14},\Lambda^2_{34},\Lambda^3_{25},
\Lambda^4_{25},\Lambda^5_{13}, \Lambda^5_{14}, \Lambda^5_{34},  \diamondsuit^5_{134}, \diamondsuit^2_{134}, \Theta_{12},\Theta_{15},\Theta_{23}, \Theta_{24},
\Theta_{35}, \Theta_{45}, \bigcirc\}$.
\end{itemize}

Then
\begin{itemize}
\item $\bar\sigma^{-1}(x)\setminus\lambda(\mathrm{M\mkern 1mu}_{3,3}^{\cdot2})=\emptyset$ for $x\in\{a^2,a^4,a^5\}$;
\item $\bar\sigma^{-1}(a^3)\setminus\lambda(\mathrm{M\mkern 1mu}_{3,3}^{\cdot2})=
    \{\triangle_{125},\Lambda^2_{15},\Lambda^2_{35}, \Lambda^2_{45},\Lambda^5_{12},\Lambda^5_{23},\Lambda^5_{24},\diamondsuit^2_{135},
    \diamondsuit^2_{145},\diamondsuit^5_{123},\diamondsuit^5_{124},\Theta_{25},\Lambda^2,\Lambda^5\}$;
\item $\bar\sigma^{-1}(\triangle_{245})\setminus\lambda(\mathrm{M\mkern 1mu}_{3,3}^{\cdot2})=\{\triangle_{134}\}$;
\item $\bar\sigma^{-1}(\triangle_{234})\setminus\lambda(\mathrm{M\mkern 1mu}_{3,3}^{\cdot2})=\{\triangle_{123},\triangle_{135},\Lambda^1_{23},\Lambda^1_{35},\Lambda^3_{12},\Lambda^3_{15},
\diamondsuit^1_{235},\diamondsuit^3_{125}, \Theta_{13}\}$;
\item $\bar\sigma^{-1}(\triangle_{235})\setminus\lambda(\mathrm{M\mkern 1mu}_{3,3}^{\cdot2})=\{\triangle_{124},\triangle_{145}, \Lambda^1_{24}, \Lambda^1_{45}, \Lambda^4_{12},\Lambda^4_{15}, \diamondsuit^1_{245},
\diamondsuit^4_{125}, \Theta_{14}\}$;
\item $\bar\sigma^{-1}(\triangle_{345})\setminus\lambda(\mathrm{M\mkern 1mu}_{3,3}^{\cdot2})=\{\Lambda^3_{24},\Lambda^3_{45},\Lambda^4_{23}, \Lambda^4_{35},\Theta_{34}\}$;
\item $\bar\sigma^{-1}(\diamondsuit^2_{345})\setminus\lambda(\mathrm{M\mkern 1mu}_{3,3}^{\cdot2})= \{\Lambda^1_{34}, \diamondsuit^1_{234},
\diamondsuit^1_{345},\Lambda^1\}$;
\item $\bar\sigma^{-1}(\diamondsuit^4_{235})\setminus\lambda(\mathrm{M\mkern 1mu}_{3,3}^{\cdot2})=\{\Lambda^3_{14},\diamondsuit^3_{124}, \diamondsuit^3_{145},\Lambda^3\}$;
\item $\bar\sigma^{-1}(\diamondsuit^5_{234})\setminus\lambda(\mathrm{M\mkern 1mu}_{3,3}^{\cdot2})=\{\Lambda^4_{13},\diamondsuit^4_{123}, \diamondsuit^4_{135},\Lambda^4\}$;

\item $\bar\sigma^{-1}(\diamondsuit^3_{245})\setminus\lambda(\mathrm{M\mkern 1mu}_{3,3}^{\cdot2})=
    \{\Lambda^1_{25},\Lambda^2_{13},\Lambda^2_{14},\Lambda^2_{34},\Lambda^3_{25}, \Lambda^4_{25},\Lambda^5_{13}, \Lambda^5_{14}, \Lambda^5_{34},
    \diamondsuit^5_{134}, \diamondsuit^2_{134}, \Theta_{12},\Theta_{15},\Theta_{23}, \Theta_{24}, \Theta_{35}, \Theta_{45}, \bigcirc\}$.
\end{itemize}

By Corollary~\ref{c:Ker}, for any automorphism $\psi$ of $\lambda(\mathrm{M\mkern 1mu}_{3,3})$ we get $|\bar\sigma^{-1}(\psi(x))|=|\bar\sigma^{-1}(x)|$ for all
$x\in\lambda(\mathrm{M\mkern 1mu}_{3,3}^{\cdot 2})$. Taking into account Corollary~\ref{c:Ker}, Claim~\ref{cl:M33} and comparing the cardinalities of
the fibers of $\bar\sigma$, we conclude that $\psi$ is the identity on the set $\lambda(\mathrm{M\mkern 1mu}_{3,3}^{\cdot2})\setminus\{\diamondsuit^2_{345},\diamondsuit^5_{234}\}$.

To see that $\psi|\lambda(\mathrm{M\mkern 1mu}_{3,3}^{\cdot2})$ is trivial, it is sufficient to show that the assumption $\psi(\diamondsuit^2_{345})=\diamondsuit^5_{234}$ leads to a contradiction. In this case $\psi(\Lambda^1)\in \{\Lambda^4_{13},\diamondsuit^4_{123}, \diamondsuit^4_{135},\Lambda^4\}$.
Replacing $\psi$ by the composition of $\psi$ with a suitable permutation in the kernel of the restriction
operator $R:\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{3,3}))\to\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{3,3}^{\cdot2}))$,
we can assume that $\psi(\Lambda^1)=\Lambda^4$ and $\psi(\triangle_{123})=\triangle_{123}$.

Now observe that $\triangle_{123}*\Lambda^1=\triangle_{234}\ne \triangle_{345}=\triangle_{123}*\Lambda^4.$
 On the other hand,
$$\triangle_{234}=\psi(\triangle_{234})=\psi(\triangle_{123}*\Lambda^1)=\psi(\triangle_{123})*\psi(\Lambda^1)=
\triangle_{123}*\Lambda^4=\triangle_{345},$$which is a desired
contradiction showing that the restriction $\psi|\lambda(\mathrm{M\mkern 1mu}_{3,3}^{\cdot2})$ is trivial.

Therefore, the restriction operator $R$ has trivial range and the automorphism
group $\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{3,3}))$ coincides with the kernel of $R$, which is isomorphic to
$$\prod_{\mathcal L\in\lambda(\mathrm{M\mkern 1mu}_{3,3}^{\cdot2})}S_{\bar\sigma^{-1}(\mathcal L)\setminus\lambda(\mathrm{M\mkern 1mu}_{3,3}^{\cdot2})}\cong
S_4^3\times S_{5}\times S_9^2\times S_{14}\times S_{18}.$$


\subsection{The semigroup $\lambda(\mathrm{M\mkern 1mu}_{4,2})$}

In this section we recognize the structure of the automorphism group of the superextension of the semigroup $\mathrm{M\mkern 1mu}_{4,2}$ and its
characteristic ideals $\mathrm{M\mkern 1mu}_{4,2}^{\cdot k}$ for $k\in\{2,3,4\}$. The monogenic semigroup $\mathrm{M\mkern 1mu}_{4,2}=\{a,a^2,a^3,a^4,a^5\}$
is generated by an element $a$ such that $a^6=a^4$.
The characteristic ideal $\mathrm{M\mkern 1mu}_{4,2}^{\cdot4}=\{a^4,a^5\}$ is a group with neutral element $a^4$.

Observe that $\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot 4})=\mathrm{M\mkern 1mu}_{4,2}^{\cdot4}\cong C_2$ and hence $\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot4}))\cong\mathrm{Aut\mkern 2mu}(C_2)\cong C_1$.

To recognize the structure of the automorphism groups of the superextensions of the semigroups $\mathrm{M\mkern 1mu}_{4,2}^{\cdot3}=\{a^3,a^4,a^5\}$
and $\mathrm{M\mkern 1mu}_{4,2}^{\cdot2}=\{a^2,a^3,a^4,a^5\}$, observe that the map $\rho:\mathrm{M\mkern 1mu}_{4,2}^{\cdot 2}\to \mathrm{M\mkern 1mu}_{4,2}^{\cdot4}$, $\rho:x\mapsto a^4x=a^2x$,
is a homomorphic retraction of $\mathrm{M\mkern 1mu}_{4,2}^{\cdot 2}$ onto the group $\mathrm{M\mkern 1mu}_{4,2}^{\cdot4}$ such that $xy=\rho(x)\cdot\rho(y)$ for
all $x,y\in\mathrm{M\mkern 1mu}_{4,2}^{\cdot2}$. This homomorphic retraction induces a homomorphic retraction
$$\bar \rho:\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot 2})\to\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot4}),\;\;\bar\rho:\mathcal A\mapsto a^4*\mathcal A,$$
such that $\mathcal A*\mathcal B=\bar\rho(\mathcal A)*\bar\rho(\mathcal B)$ for any $\mathcal A,\mathcal B\in\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot2})$.

Using the notations for the elements of the superextension $\lambda(\mathrm{M\mkern 1mu}_{4,2})$ from Subsection~\ref{ss:M24}, observe that $$\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot2})=\{a^i:2\le i\le 5\}\cup\{\triangle_{ijk}:2\le i<j<k\le 5\}\cup\big\{\diamondsuit^n_{ijk}:2\le i<j<k\le 5,\;n\in\{2,3,4,5\}\setminus\{i,j,k\}\big\}.$$
By routine calculations it can be shown that the map $\bar\rho:\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot2})\to\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot4})=\{a^4,a^5\}$ has the following fibers:
\begin{itemize}
\item $\bar\rho^{-1}(a^4)=\{a^2,a^4,\triangle_{234},\triangle_{245}, \diamondsuit^2_{345},\diamondsuit^4_{235}\}$;
\item $\bar\rho^{-1}(a^5)=\{a^3,a^5,\triangle_{235},\triangle_{345}, \diamondsuit^3_{245},\diamondsuit^5_{234}\}$.
\end{itemize}
Now we see that $\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot3}))$ is isomorphic to $$\prod_{i\in\{4,5\}}S_{\bar\rho^{-1}(a^i)\cap\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot3}) \setminus\mathrm{M\mkern 1mu}_{4,2}^{\cdot4}}=S_{\{a^3,\triangle_{345}\}}\cong C_2$$
and $\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot2}))$ is isomorphic to
$$\prod_{i\in\{4,5\}}S_{\bar\rho^{-1}(a^i)\setminus\mathrm{M\mkern 1mu}_{4,2}^{\cdot4}}\cong
S_5\times S_5.$$

To detect the algebraic structure of the automorphism group of $\lambda(\mathrm{M\mkern 1mu}_{4,2})$, consider the map $\sigma:\mathrm{M\mkern 1mu}_{4,2}\to\mathrm{M\mkern 1mu}_{4,2}^{\cdot2}$, $\sigma:x\mapsto ax$, which induces the map
$$\bar\sigma:\lambda(\mathrm{M\mkern 1mu}_{4,2})\to\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot2}),\;\;\bar\sigma:\mathcal A\mapsto a*\mathcal A.$$
By Lemma~\ref{l:auto-shift}, $\bar\sigma$ is an auto-shift of $\lambda(\mathrm{M\mkern 1mu}_{4,2})$.

By routine calculations we can show that the map $\bar\sigma$ has the following fibers:
\begin{itemize}
\item $\bar\sigma^{-1}(a^2)=\{a\}$, $\bar\sigma^{-1}(a^3)=\{a^2\}$, $\bar\sigma^{-1}(a^5)=\{a^4\}$;
\item $\bar\sigma^{-1}(a^4)=\{a^3,a^5,\triangle_{135},\triangle_{235}, \triangle_{345},\Lambda^3_{15},\Lambda^3_{25}, \Lambda^3_{45},
\Lambda^5_{13}, \Lambda^5_{23}, \Lambda^5_{34},\diamondsuit^3_{125}, \diamondsuit^3_{145}, \diamondsuit^3_{245},\diamondsuit^5_{123},
\diamondsuit^5_{134}, \diamondsuit^5_{234}, \Lambda^3,\Lambda^5, \Theta_{35}\}$;
\item $\bar\sigma^{-1}(\triangle_{235})=\{\triangle_{124}\}$;
\item $\bar\sigma^{-1}(\triangle_{234})=\{\triangle_{123},\triangle_{125}, \Lambda^1_{23}, \Lambda^1_{25}, \Lambda^2_{13},\Lambda^2_{15}, \diamondsuit^1_{235}, \diamondsuit^2_{135}, \Theta_{12}\}$;
\item $\bar\sigma^{-1}(\triangle_{245})=\{\triangle_{134},\triangle_{145}, \Lambda^1_{34},\Lambda^1_{45},
\Lambda^4_{13},\Lambda^4_{15}, \diamondsuit^1_{345}, \diamondsuit^4_{135},\Theta_{14}\}$;
\item $\bar\sigma^{-1}(\triangle_{345})=\{\triangle_{234},\triangle_{245}, \Lambda^2_{34}, \Lambda^2_{45}, \Lambda^4_{23}, \Lambda^4_{25},
\diamondsuit^2_{345}, \diamondsuit^4_{235},\Theta_{24}\}$;
\item $\bar\sigma^{-1}(\diamondsuit^2_{345}) = \{\diamondsuit^1_{234},\diamondsuit^1_{245},\Lambda^1, \Lambda^1_{24}\}$;
\item $\bar\sigma^{-1}(\diamondsuit^3_{245})= \{\diamondsuit^2_{134}, \diamondsuit^2_{145}, \Lambda^2, \Lambda^2_{14}\}$;
\item $\bar\sigma^{-1}(\diamondsuit^5_{234})= \{\diamondsuit^4_{123},\diamondsuit^4_{125},\Lambda^4, \Lambda^4_{12}\}$;
\item $\bar\sigma^{-1}(\diamondsuit^4_{235})= \{\Lambda^1_{35},\Lambda^2_{35},\Lambda^3_{12},\Lambda^3_{14},
\Lambda^3_{24},\Lambda^4_{35},\Lambda^5_{12},\Lambda^5_{14},\Lambda^5_{24}, \diamondsuit^3_{124},\diamondsuit^5_{124},
\Theta_{13},\Theta_{15},\Theta_{23}, \Theta_{25},\Theta_{34},\Theta_{45},\bigcirc\}$.
\end{itemize}

Now consider the restriction operator $$R:\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{4,2}))\to\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot2})),\;\;R:\psi\mapsto\psi|\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot2}).$$
 We claim that this operator has trivial range.


Fix any automorphism $\psi\in\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{4,2}))$. Proposition~\ref{mono1} ensures that $\psi(a^i)=a^i$ for all $i\in\{1,2,3,4,5\}$.


The description of the automorphism group $\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot2}))$ ensures that the sets $\{a^2,a^4,\triangle_{234},\triangle_{245}, \diamondsuit^2_{345},\diamondsuit^4_{235}\}$ and $\{a^3,a^5,\triangle_{235},\triangle_{345}, \diamondsuit^3_{245},\diamondsuit^5_{234}\}$ are $\psi$-invariant.
Then the sets $F_4=\{\triangle_{234},\triangle_{245}, \diamondsuit^2_{345},\diamondsuit^4_{235}\}$
and $F_4'=\{\triangle_{235},\triangle_{345}, \diamondsuit^3_{245},\diamondsuit^5_{234}\}$ are $\psi$-invariant, too.
Taking into account that the ideal $\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot3})=\{a^3,a^4,a^5,\triangle_{345}\}$ is characteristic,
we conclude that $\psi(\triangle_{345})=\triangle_{345}$. Consequently, the set
$F_3=\{\triangle_{235}, \diamondsuit^3_{245},\diamondsuit^5_{234}\}$ is $\psi$-invariant and $\psi(\mathcal A)=\mathcal A$ for any
$\mathcal A\in\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot2})\setminus(F_3\cup F_4)$.



It follows from $\psi\circ\bar\sigma=\bar\sigma\circ\psi$ that $$|\bar\sigma^{-1}(\psi^{-1}(\mathcal A))\setminus\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot2})|=
|\bar\sigma^{-1}(\mathcal A)\setminus\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot2})|$$ for any $\mathcal A\in\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot2})$.
Observe that
\begin{itemize}
\item $|\bar\sigma^{-1}(x)\setminus\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot2})|=0$ for $x\in\{a^2,a^3,a^5\}$;

\item $|\bar\sigma^{-1}(a^4)\setminus\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot2})|= |\{\triangle_{135},\Lambda^3_{15},\Lambda^3_{25}, \Lambda^3_{45},\Lambda^5_{13},
\Lambda^5_{23}, \Lambda^5_{34},\diamondsuit^3_{125}, \diamondsuit^3_{145}, \diamondsuit^5_{123},\diamondsuit^5_{134}, \Lambda^3,\Lambda^5, \Theta_{35}\}|=14$;
\item $|\bar\sigma^{-1}(\triangle_{235})\setminus\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot2})|=|\{\triangle_{124}\}|=1$;
\item $|\bar\sigma^{-1}(\triangle_{234})\setminus\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot2})|=
   |\{\triangle_{123},\triangle_{125}, \Lambda^1_{23}, \Lambda^1_{25}, \Lambda^2_{13},\Lambda^2_{15}, \diamondsuit^1_{235}, \diamondsuit^2_{135}, \Theta_{12}\}|=9$;
\item  $|\bar\sigma^{-1}(\triangle_{245})\setminus\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot2})|=
|\{\triangle_{134},\triangle_{145}, \Lambda^1_{34},\Lambda^1_{45}, \Lambda^4_{13},\Lambda^4_{15},
\diamondsuit^1_{345},\diamondsuit^4_{135},\Theta_{14}\}|=9$;
\item  $|\bar\sigma^{-1}(\triangle_{345})\setminus\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot2})|=
|\{\Lambda^2_{34}, \Lambda^2_{45}, \Lambda^4_{23}, \Lambda^4_{25},  \Theta_{24}\}|=5$;
\item $|\bar\sigma^{-1}(\diamondsuit^2_{345})\setminus\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot2})|=
|\{\diamondsuit^1_{234},\diamondsuit^1_{245},\Lambda^1,\Lambda^1_{24}\}|=4$;
\item $|\bar\sigma^{-1}(\diamondsuit^3_{245})\setminus\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot2})|=
|\{\diamondsuit^2_{134},\diamondsuit^2_{145}, \Lambda^2,  \Lambda^2_{14}\}|=4$;
\item $|\bar\sigma^{-1}(\diamondsuit^5_{234})\setminus\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot2})|=
|\{\diamondsuit^4_{123},\diamondsuit^4_{125},\Lambda^4, \Lambda^4_{12}\}|=4$;
\item  $|\bar\sigma^{-1}(\diamondsuit^4_{235})\setminus\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot2})|=$\newline $=|\{\Lambda^1_{35},\Lambda^2_{35},\Lambda^3_{12},
\Lambda^3_{14}, \Lambda^3_{24},\Lambda^4_{35},\Lambda^5_{12},\Lambda^5_{14},\Lambda^5_{24}, \diamondsuit^3_{124},\diamondsuit^5_{124},
\Theta_{13},\Theta_{15},\Theta_{23}, \Theta_{25},\Theta_{34},\Theta_{45},\bigcirc\}|=18$.

\end{itemize}
Comparing the cardinalities of the sets $\bar\sigma^{-1}(\mathcal A)\setminus\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot2})$ for points in the sets $F_3$ and $F_4$,
we can conclude that the sets $\{\triangle_{234},\triangle_{245}\}$ and $\{\diamondsuit^3_{245},\diamondsuit^5_{234}\}$ are $\psi$-invariant and $\psi(\mathcal A)=\mathcal A$ for any $\mathcal A\in \lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot2})\setminus
\{\triangle_{234},\triangle_{245},\diamondsuit^3_{245},\diamondsuit^5_{234}\}$.

Now we show that $\psi(\triangle_{234})=\triangle_{234}$. In the opposite case, $\psi(\triangle_{234})=\triangle_{245}$ and hence $\psi(\triangle_{123})\in\bar\sigma^{-1}(\triangle_{245})\setminus\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot2})$.
Replacing $\psi$ by the composition of $\psi$ with a suitable permutation in the kernel of the restriction operator $R:\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{4,2}))\to\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot2}))$, we can assume that $\psi(\triangle_{123})=\triangle_{134}$.

Taking into account that $\triangle_{123}*\triangle_{123}=\diamondsuit^4_{235}$ and $\triangle_{134}*\triangle_{134}=\triangle_{245}$, we conclude that
$$\diamondsuit^4_{235}=\psi(\diamondsuit^4_{235})=\psi(\triangle_{123})*
\psi(\triangle_{123})=\triangle_{134}*\triangle_{134}=\triangle_{245},$$which is a contradiction proving that $\psi(\mathcal A)=\mathcal A$ for all $\mathcal A\in \lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot2})\setminus
\{\diamondsuit^3_{245},\diamondsuit^5_{234}\}$.

Assuming that $\psi|\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot2})$ is not identity, we conclude that $\psi(\diamondsuit^3_{245})=\diamondsuit^5_{234}$. Replacing $\psi$ by the composition of $\psi$ with a suitable permutation in the kernel of the restriction operator $R$, we can assume that $\psi(\Lambda^2)=\Lambda^4$ and $\psi(\triangle_{123})=\triangle_{123}$.

Taking into account that $\triangle_{123}*\Lambda^2=\triangle_{345}$ and $\triangle_{123}*\Lambda^4=\diamondsuit^5_{234}$, we conclude that
$$\triangle_{345}=\psi(\triangle_{345})=\psi(\triangle_{123}*\Lambda^2)= \psi(\triangle_{123})*\psi(\Lambda^2)=\triangle_{123}*\Lambda^4=\diamondsuit^5_{234},$$
which is a contradiction completing the proof of the triviality of the range of the restriction operator $R:\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{4,2}))\to\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot2}))$.

Then the automorphism group
$\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{4,2}))$ coincides with the kernel of $R$, which is isomorphic to the group
$$\prod_{\mathcal L\in\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot2})}S_{\bar\sigma^{-1}(\mathcal L)\setminus\lambda(\mathrm{M\mkern 1mu}_{4,2}^{\cdot2})}\cong
S_4^3\times S_5\times S_9^2\times S_{14}\times S_{18}.$$



\subsection{The semigroup $\lambda(\mathrm{M\mkern 1mu}_{5,1})$}

In this section we recognize the structure of the automorphism group of the superextension of the semigroup $\mathrm{M\mkern 1mu}_{5,1}$ and its
characteristic ideals $\mathrm{M\mkern 1mu}_{5,1}^{\cdot k}$ for $k\in\{2,3,4,5\}$. The monogenic semigroup $\mathrm{M\mkern 1mu}_{5,1}=\{a,a^2,a^3,a^4,a^5\}$ is
generated by an element $a$ such that $a^6=a^5$.
Observe that for any $x, y\in\{a^3,a^4,a^5\}=\mathrm{M\mkern 1mu}_{5,1}^{\cdot3}$ we get $xy=a^5$. This implies that for any $k\in\{3,4,5\}$ any
bijection $\psi$ of $\lambda(\mathrm{M\mkern 1mu}_{5,1}^{\cdot k})$ with $\psi(a^5)=a^5$ is an automorphism of the semigroup $\lambda(\mathrm{M\mkern 1mu}_{5,1}^{\cdot k})$.
This observation implies that $\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{5,1}^{\cdot5}))\cong \mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{5,1}^{\cdot4}))\cong C_1$ and $\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{5,1}^{\cdot 3}))\cong S_3$.

Next, consider the semigroup $\mathrm{M\mkern 1mu}_{5,1}^{\cdot2}=\{a^2,a^3,a^4,a^5\}$ and its superextension $\lambda(\mathrm{M\mkern 1mu}_{5,1}^{\cdot2})$. It follows from $\mathrm{M\mkern 1mu}_{5,1}^{\cdot2}*\mathrm{M\mkern 1mu}_{5,1}^{\cdot2}=\mathrm{M\mkern 1mu}_{5,1}^{\cdot 4}=\{a^4,a^5\}$ that $\lambda(\mathrm{M\mkern 1mu}_{5,1}^{\cdot2})*\lambda(\mathrm{M\mkern 1mu}_{5,1}^{\cdot2})=\lambda(\mathrm{M\mkern 1mu}_{5,1}^{\cdot4})=\mathrm{M\mkern 1mu}_{5,1}^{\cdot4}$.
This implies that the ideal $\mathrm{M\mkern 1mu}_{5,1}^{\cdot4}$ is characteristic in the semigroup $\lambda(\mathrm{M\mkern 1mu}_{5,1}^{\cdot2})$. Since the semigroup $\mathrm{M\mkern 1mu}_{5,1}^{\cdot4}=\{a^4,a^5\}$ has trivial automorphism group, $\psi(a^4)=a^4$ and $\psi(a^5)=a^5$ for any automorphism $\psi$ of $\lambda(\mathrm{M\mkern 1mu}_{5,1}^{\cdot2})$.

It is easy to show that the equation $x*y=a^4$ has a unique solution $x=y=a^2$ in the semigroup $\lambda(\mathrm{M\mkern 1mu}_{5,1}^{\cdot2})$.
Consequently, $\psi(a^2)=a^2$ for any automorphism $\psi$ of $\lambda(\mathrm{M\mkern 1mu}_{5,1}^{\cdot2})$ and $\mathcal A*\mathcal B=a^5$ for any maximal
linked upfamilies $\mathcal A,\mathcal B\in\lambda(\mathrm{M\mkern 1mu}_{5,1}^{\cdot2})$ such that $(\mathcal A,\mathcal B)\ne(a^2,a^2)$. It follows that any bijection
$\psi$ of $\lambda(\mathrm{M\mkern 1mu}_{5,1}^{\cdot2})$ such that $\psi(a^i)=a^i$ for $i\in\{2,4,5\}$ is an automorphism of the semigroup
$\lambda(\mathrm{M\mkern 1mu}_{5,1}^{\cdot 2})$. This implies that $\mathrm{Aut\mkern 2mu}(\mathrm{M\mkern 1mu}_{5,1}^{\cdot2})\cong S_9$.

To recognize the automorphism group of $\lambda(\mathrm{M\mkern 1mu}_{5,1})$, consider
the map $$\bar\sigma:\lambda(\mathrm{M\mkern 1mu}_{5,1})\to\lambda(\mathrm{M\mkern 1mu}_{5,1}^{\cdot2}),\;\;\bar\sigma:\mathcal A\mapsto a*\mathcal A.$$
By Lemma~\ref{l:auto-shift}, $\bar\sigma$ is an auto-shift of $\lambda(\mathrm{M\mkern 1mu}_{5,1})$ and its restriction
$\bar\sigma_2:=\bar\sigma|\lambda(\mathrm{M\mkern 1mu}_{5,1}^{\cdot2})$ is a good shift of $\lambda(\mathrm{M\mkern 1mu}_{5,1}^{\cdot2})$.

By routine calculations, it can be shown that the map $\bar\sigma$ has the following fibers:
\begin{itemize}
\item $\bar\sigma^{-1}(a^{i})=\{a^{i-1}\}$ for $i\in\{2,3,4\}$;
\item $\bar\sigma^{-1}(a^5)=\{a^4,a^5,\Lambda^5,\Lambda^4,\diamondsuit^5_{234}, \diamondsuit^5_{134},\diamondsuit^5_{124},\diamondsuit^4_{235},
\diamondsuit^4_{135}, \diamondsuit^4_{125},\Lambda^5_{34},\Lambda^5_{24},\Lambda^5_{14},\Lambda^4_{35}, \Lambda^4_{25},\Lambda^4_{15},
\triangle_{345},\triangle_{245}, \triangle_{145}, \Theta_{45}\}$;
\item $\bar\sigma^{-1}(\triangle_{234})=\{\triangle_{123}\}$;
\item $\bar\sigma^{-1}(\triangle_{235})=\{\Theta_{12},\diamondsuit^2_{145}, \diamondsuit^1_{245},\Lambda^2_{15},\Lambda^2_{14},\Lambda^1_{25},\Lambda^1_{24}, \triangle_{125},\triangle_{124}\}$;
\item $\bar\sigma^{-1}(\triangle_{245})=\{\Theta_{13},\diamondsuit^3_{145}, \diamondsuit^1_{345},\Lambda^3_{15},\Lambda^3_{14},\Lambda^1_{35},\Lambda^1_{34}, \triangle_{135},\triangle_{134}\}$;
\item $\bar\sigma^{-1}(\triangle_{345})=\{\Theta_{23},\diamondsuit^3_{245}, \diamondsuit^2_{345},\Lambda^3_{25},\Lambda^3_{24},\Lambda^2_{35},\Lambda^2_{34}, \triangle_{234},\triangle_{235}\}$;
\item $\bar\sigma^{-1}(\diamondsuit^2_{345})=\{\Lambda^1,\diamondsuit^1_{235}, \diamondsuit^1_{234},\Lambda^1_{23}\}$;
\item $\bar\sigma^{-1}(\diamondsuit^3_{245})=\{\Lambda^2,\diamondsuit^2_{135}, \diamondsuit^2_{134},\Lambda^2_{13}\}$;
\item $\bar\sigma^{-1}(\diamondsuit^4_{235})=\{\Lambda^3,\diamondsuit^3_{125}, \diamondsuit^3_{124},\Lambda^3_{12}\}$;
\item $\bar\sigma^{-1}(\diamondsuit^5_{234})=\{\bigcirc,\Theta_{35}, \Theta_{34},\Theta_{25},\Theta_{24},\Theta_{15},\Theta_{14},
\diamondsuit^5_{123}, \diamondsuit^4_{123},\Lambda^5_{23},\Lambda^5_{13},\Lambda^5_{12},\Lambda^4_{23},\Lambda^4_{13}, \Lambda^4_{12},\Lambda^3_{45},
\Lambda^2_{45}, \Lambda^1_{45}\}$.
\end{itemize}

To recognize the algebraic structure of the automorphism group $\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{5,1}))$, consider the restriction operator
$R:\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{5,1}))\to\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{5,1}^{\cdot2}))$. By Corollary~\ref{c:Ker}, the kernel of this operator is isomorphic to
$$\prod_{\mathcal L\in\lambda(\mathrm{M\mkern 1mu}_{5,1}^{\cdot2})}S_{\bar\sigma^{-1}(\mathcal L)\setminus \lambda(\mathrm{M\mkern 1mu}_{5,1}^{\cdot2})}\cong
S_{14}\times S_9^2\times S_5\times S_4^3\times S_{18}.$$

We claim that the operator $R$ has trivial range. Given any automorphism $\psi$ of $\lambda(\mathrm{M\mkern 1mu}_{5,1})$, we should prove that $\psi(\mathcal A)=\mathcal A$ for any $\mathcal A\in\lambda(\mathrm{M\mkern 1mu}_{5,1}^{\cdot2})$. By Proposition~\ref{mono1}, $\psi(a^i)=a^i$ for all $i\in\{1,2,3,4,5\}$. The equality $\psi(a)=a$ implies that $\bar\sigma\circ\psi=\psi\circ\bar\sigma$. Consequently, for every $\mathcal L\in \lambda(\mathrm{M\mkern 1mu}_{5,1}^{\cdot2})$ we have $\psi(\bar\sigma^{-1}(\mathcal L))=\bar\sigma^{-1}(\psi(\mathcal L))$. Since the ideals $\lambda(\mathrm{M\mkern 1mu}_{5,1}^{\cdot k})$ are characteristic in $\lambda(\mathrm{M\mkern 1mu}_{5,1})$, Corollary~\ref{c:Ker} ensures that
   $$|\bar\sigma^{-1}(\psi(\mathcal L))\cap\lambda(\mathrm{M\mkern 1mu}_{5,1}^{\cdot k})|=|\bar\sigma^{-1}(\mathcal L)\cap\lambda(\mathrm{M\mkern 1mu}_{5,1}^{\cdot k})|$$ for all
$\mathcal L\in\lambda(\mathrm{M\mkern 1mu}_{5,1}^{\cdot 2})$ and all $k\in\{1,2,3,4,5\}$.



Comparing the cardinalities of the sets $\bar\sigma^{-1}(\mathcal L)\cap\lambda(\mathrm{M\mkern 1mu}_{5,1}^{\cdot k})$ for various $k\in\{1,2,3,4,5\}$ and $\mathcal L\in\lambda(\mathrm{M\mkern 1mu}_{5,1}^{\cdot 2})$, we see that $\psi(F_2)=F_2$, $\psi(F_3)=F_3$ and $\psi(\mathcal A)=\mathcal A$ for all $\mathcal A\in\mathrm{M\mkern 1mu}_{5,1}^{\cdot2}\setminus (F_2\cup F_3)$, where  $F_2=\{\triangle_{235},\triangle_{245}\}$ and $F_3=\{\diamondsuit^2_{345},\diamondsuit^3_{245},\diamondsuit^4_{235}\}$.

Let us check that $\psi(\triangle_{235})=\triangle_{235}$. To derive a contradiction, assume that $\psi(\triangle_{235})\ne\triangle_{235}$ and hence $\psi(\triangle_{235})=\triangle_{245}$. Consider the element $\triangle_{124}\in \bar\sigma^{-1}(\triangle_{235})\setminus\lambda(\mathrm{M\mkern 1mu}_{5,1}^{\cdot2})$ and observe that $\psi(\triangle_{124})\in\psi(\bar\sigma^{-1}(\triangle_{235}))= \bar\sigma^{-1}(\psi(\triangle_{235}))= \bar\sigma^{-1}(\triangle_{245})$.
Let $\pi:\lambda(\mathrm{M\mkern 1mu}_{5,1})\to\lambda(\mathrm{M\mkern 1mu}_{5,1})$ be a permutation such that $\pi(\psi(\triangle_{124}))=\triangle_{135}$ and $\pi(\mathcal A)=\mathcal A$ for all $\mathcal A\in\lambda(\mathrm{M\mkern 1mu}_{5,1})\setminus\{\triangle_{135},\triangle_{124}\}$.
By Corollary~\ref{c:Ker}, the permutation $\pi$ belongs to the automorphism group $\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{5,1}))$. Replacing $\psi$ by $\pi\circ\psi$, we can assume that $\psi(\triangle_{124})=\triangle_{135}$.

Observe that $$a^5=\triangle_{135}*\triangle_{123}=
\psi(\triangle_{124})*\psi(\triangle_{123})=\psi(\triangle_{124}*\triangle_{123})=
\psi(\triangle_{345})\ne a^5.$$
This contradiction shows that $\psi(\triangle_{235})=\triangle_{235}$ and hence $\psi(\triangle_{245})=\triangle_{245}$ (as $\psi(F_2)=F_2)$).

Next, we show that $\psi(\mathcal A)=\mathcal A$ for any $\mathcal A\in F_3=\{\diamondsuit^2_{345},\diamondsuit^3_{245},\diamondsuit^4_{235}\}$.
To simplify notations, put $\diamondsuit^2:=\diamondsuit^2_{345}$,
$\diamondsuit^3:=\diamondsuit^3_{245}$, $\diamondsuit^4:=\diamondsuit^4_{235}$.
To derive a contradiction, assume that $\psi(\diamondsuit^i)=\diamondsuit^j$ for some $2\le i<j\le 4$.
Consider the maximal linked upfamily $\Lambda^{i-1}\in \bar\sigma^{-1}(\diamondsuit^i)$ and
observe that $\psi(\Lambda^{i-1})\in\psi(\bar\sigma^{-1}(\diamondsuit^{i}))=
\bar\sigma^{-1}(\psi(\diamondsuit^i))=\bar\sigma^{-1}(\diamondsuit^{j})$.
Since $\Lambda^{j-1}\in \bar\sigma^{-1}(\diamondsuit^{j})$, we can replace $\psi$ by the
composition with the permutation exchanging $\psi(\Lambda^{i-1})$ with $\Lambda^{j-1}$
and $\psi(\triangle_{134})$ with $\triangle_{134}$, and assume that $\psi(\Lambda^{i-1})=\Lambda^{j-1}$ and $\psi(\triangle_{134})=\triangle_{134}$.

 Now observe that the elements $\Lambda^1,\Lambda^2,\Lambda^3$ can be algebraically distinguished by the equalities:
$$
\begin{gathered}
\Lambda^1*\triangle_{123}=\triangle_{345},\;\;\Lambda^1*\triangle_{134}=\triangle_{345}\\
\Lambda^2*\triangle_{123}=\triangle_{345},\;\;\Lambda^2*\triangle_{134}=a^5\\
\Lambda^3*\triangle_{123}=a^5,\;\;\Lambda^3*\triangle_{134}=a^5.
\end{gathered}
$$
Two cases are possible. If $i=2$, then $\psi(\Lambda^1)=\Lambda^{j-1}$ and we obtain a contradiction: $$a^5=\psi(a^5)=\Lambda^{j-1}*\triangle_{134}=\psi(\Lambda^1)*\psi(\triangle_{134})=
\psi(\Lambda^1*\triangle_{134})=\psi(\triangle_{345})\ne a^5.$$
If $i=3$, then we obtain a contradiction considering
$$a^5=\psi(a^5)=\Lambda^{j-1}*\triangle_{123}=\psi(\Lambda^2)*\psi(\triangle_{123})=
\psi(\Lambda^2*\triangle_{123})=\psi(\triangle_{345})\ne a^5.$$
In both cases we obtain a contradiction with the assumption that $\psi(\mathcal A)\ne\mathcal A$ for some $\mathcal A\in F_3$. This contradiction completes the proof of the triviality of the range of the operator $R$. Then the group $\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{5,1}))$ is equal to the kernel of the operator $R$ and hence is isomorphic to the group
$
S_4^3\times S_5\times S_9^2\times S_{14}\times S_{18}$.
\newpage

\section{Summary Table and some Conjectures}\label{s:table}

The obtained results on the automorphism groups of superextensions of monogenic semigroups of cardinality $\le 5$ are summed up in the following table.

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$S$ & $|S|=$& $S\cong$& $\mathrm{Aut\mkern 2mu}(S)\cong$&$\mathrm{Aut\mkern 2mu}(\lambda(S))\cong$&$|\mathrm{Aut\mkern 2mu}(\lambda(S))|=$\\
\hline
$\mathrm{M\mkern 1mu}_{1,1}$ &1&$C_1$&$C_1$&$C_1$&1\\
\hline
$\mathrm{M\mkern 1mu}_{1,2}$ &2&$C_2$&$C_1$&$C_1$&1\\
\hline
$\mathrm{M\mkern 1mu}_{2,1}$ &2&  &$C_1$&$C_1$&1\\
$\mathrm{M\mkern 1mu}_{2,1}^{\cdot2}$ &1& $C_1$&$C_1$&$C_1$&1\\
\hline
$\mathrm{M\mkern 1mu}_{1,3}$ &3& $C_3$&$C_2$&$C_2$&2\\
\hline
$\mathrm{M\mkern 1mu}_{2,2}$ &3& &$C_1$&$C_2$&2\\
$\mathrm{M\mkern 1mu}_{2,2}^{\cdot2}$ &2& $C_2$&$C_1$&$C_1$&1\\
\hline
$\mathrm{M\mkern 1mu}_{3,1}$ &3& &$C_1$&$C_1$&1\\
$\mathrm{M\mkern 1mu}_{3,1}^{\cdot2}$ &2& $\mathrm{M\mkern 1mu}_{2,1}$&$C_1$&$C_1$&1\\
$\mathrm{M\mkern 1mu}_{3,1}^{\cdot3}$ &1& $C_1$&$C_1$&$C_1$&1\\
\hline
$\mathrm{M\mkern 1mu}_{1,4}$ &4& $C_4$&$C_2$&$C_2^2$&4\\
\hline
$\mathrm{M\mkern 1mu}_{2,3}$ &4& &$C_1$&$S_3\times S_5$&720\\
$\mathrm{M\mkern 1mu}_{2,3}^{\cdot2}$ &3& $C_3$&$C_2$&$C_2$&2\\
\hline
$\mathrm{M\mkern 1mu}_{3,2}$ &4& &$C_1$&$S_3\times S_4$&144\\
$\mathrm{M\mkern 1mu}_{3,2}^{\cdot2}$ &3& $\mathrm{M\mkern 1mu}_{2,2}$&$C_1$&$C_2$&2\\
$\mathrm{M\mkern 1mu}_{3,2}^{\cdot3}$ &2& $C_2$&$C_1$&$C_1$&1\\
\hline
$\mathrm{M\mkern 1mu}_{4,1}$ &4& &$C_1$&$S_3\times S_4$&144\\
$\mathrm{M\mkern 1mu}_{4,1}^{\cdot2}$ &3&  &$C_2$&$S_3$&6\\
$\mathrm{M\mkern 1mu}_{4,1}^{\cdot3}$ &2& $\mathrm{M\mkern 1mu}_{2,1}$&$C_1$&$C_1$&1\\
$\mathrm{M\mkern 1mu}_{4,1}^{\cdot4}$ &1& $C_1$&$C_1$&$C_1$&1\\
\hline
$\mathrm{M\mkern 1mu}_{1,5}$ &5& $C_5$&$C_4$&$C_4$&4\\
\hline
$\mathrm{M\mkern 1mu}_{2,4}$ &5& &$C_1$&$S_3^3{\times} S_8^3{\times} S_{17}{\times} S_{19}$&4\,465\,152\\
$\mathrm{M\mkern 1mu}_{2,4}^{\cdot2}$ &4& $C_4$&$C_2$&$C_2^2$&4\\
\hline
$\mathrm{M\mkern 1mu}_{3,3}$ &5& &$C_1$&$S_4^3{\times} S_{5}{\times} S_9^2{\times} S_{14}{\times} S_{18}$&6\,531\,840\\
$\mathrm{M\mkern 1mu}_{3,3}^{\cdot2}$ &4& $\mathrm{M\mkern 1mu}_{2,3}$ &$C_1$&$S_3\times S_5$&720\\
$\mathrm{M\mkern 1mu}_{3,3}^{\cdot3}$ &3&$C_3$&$C_2$&$C_2$&2\\
\hline
$\mathrm{M\mkern 1mu}_{4,2}$ &5& &$C_1$&$S_4^3{\times} S_5{\times} S_9^2{\times} S_{14}{\times} S_{18}$&6\,531\,840\\
$\mathrm{M\mkern 1mu}_{4,2}^{\cdot2}$ &4& & $C_2$&$S_5\times S_5$&14400\\
$\mathrm{M\mkern 1mu}_{4,2}^{\cdot3}$ &3& & $C_1$&$C_2$&2\\
$\mathrm{M\mkern 1mu}_{4,2}^{\cdot4}$ &2&$C_2$&$C_1$&$C_1$&1\\
\hline
$\mathrm{M\mkern 1mu}_{5,1}$ &5& &$C_1$&$S_4^3{\times} S_5{\times} S_9^2{\times} S_{14}{\times} S_{18}$&6\,531\,840\\
$\mathrm{M\mkern 1mu}_{5,1}^{\cdot2}$ &4&  &$C_1$&$S_9$&362\,880\\
$\mathrm{M\mkern 1mu}_{5,1}^{\cdot3}$ &3& $\mathrm{M\mkern 1mu}_{4,1}^{\cdot2}$&$C_2$&$S_3$&6\\
$\mathrm{M\mkern 1mu}_{5,1}^{\cdot4}$ &2& $\mathrm{M\mkern 1mu}_{2,1}$&$C_1$&$C_1$&1\\
$\mathrm{M\mkern 1mu}_{5,1}^{\cdot5}$ &1& $C_1$&$C_1$&$C_1$&1\\
\hline
\end{tabular}
\end{center}
\bigskip

Analyzing the entries of this table and the arguments in Section~\ref{s4}, we can make the following conjectures.

\begin{conjecture} For any integer numbers $r,s\ge 3$ and $n,m\ge 1$ with $r+m=s+n$ the automorphism groups $\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{r,m}))$ and $\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{s,n}))$ are isomorphic.
\end{conjecture}

\begin{conjecture} For any integer $r\ge 2$ and $m\ge 1$ the restriction operator $R:\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{r,m}))\to\mathrm{Aut\mkern 2mu}(\lambda(\mathrm{M\mkern 1mu}_{r,m}^{\cdot2}))$, $R:\psi\mapsto \psi|\lambda(\mathrm{M\mkern 1mu}_{r,m}^{\cdot2})$,
has trivial range.
\end{conjecture}


