\section{Introduction}\label{intro}
A natural generalization of conjugacy is twisted conjugacy and the study of this can be dated back to $1939$ by the work of Gantmacher (see \cite{gant}). 
Let $G$ be a group and $\varphi$ an automorphism of $G$. Two elements $x, y\in G$ are said to be $\varphi$-twisted conjugate, denoted by $x\sim_{\varphi} y$, if $y=gx\varphi(g)^{-1}$ for some $g\in G$. Clearly, $\sim_{\varphi}$ is an equivalence relation on $G$. The equivalence classes with respect to this relation are called the \textit{$\varphi$-twisted conjugacy classes} or \textit{the Reidemeister classes} of $\varphi$. If $\varphi=\mathrm{Id}$, then the $\varphi$-twisted conjugacy classes are the (ordinary) conjugacy classes. The $\varphi$-twisted conjugacy class containing $x\in G$ is denoted by $[x]_{\varphi}$ or simply $[x]$, when $\varphi$ is clear from the context. The Reidemeister number of $\varphi$, denoted by $R(\varphi)$, is the number of all $\varphi$-twisted conjugacy classes. 
This number is either a natural number or $\infty$.
A group $G$ is said to have the \textit{$R_{\infty}$-property} if $R(\varphi)$ is infinite for every automorphism $\varphi$ of $G$. 

The Reidemeister number is closely related to the Nielsen number of a self-map of a manifold. More precisely, let $M$ be a closed connected manifold of dimension at least $3$ and $f: M\to M$ a continuous map, the minimal number of fixed points among all continuous maps  homotopic to $f$ is the Nielsen number $N(f)$. Note that the Nielsen number $N(f)$ is always finite and is homotopy invariant. The computation of $N(f)$ is difficult and a central object to study in the Nielsen-Reidemeister fixed point theory (cf. \cite{fel10}). This number is bounded above by the Reidemeister number $R(f_{\#})$, where $f_{\#} $ is the induced map on $ \pi_1(M) $. If $M$ is a Jiang-type space, then either $N(f)=0$ or $N(f)=R(f_{\#})$ (we refer the reader to \cite[p. 37, Theorem 5.6]{Jiang} for details). Thus, in this scenario, if 
$R(f_{\#})=\infty$ then $N(f)=0$ which in turn implies that one can deform $f$, up to homotopy, to a fixed-point free map. 

The classes of groups which have the $R_{\infty}$-property is a robust and active area of research begun by Fel'shtyn and Hill in \cite{fh94}. The problem has a long history of references, for example, see \cite{fel10, ft15} and the references therein. 
In the context of algebraic groups, Steinberg proved that
for a connected linear algebraic group $G$ over an algebraically closed field $k$ and a surjective endomorphism $\varphi$ of
G, if the fixed-point subgroup $G^{\varphi}$ is finite then $R(\varphi)=1$, and hence $G$ does not satisfy the $R_{\infty}$-property (see \cite[Theorem 10.1]{St}). The above result is classically known as \emph{Lang-Steinberg theorem}. Recently, Bose and the first author of this paper proved that a connected linear algebraic group $G$ has the (algebraic) $R_{\infty}$-property if and only if the fixed-point subgroup $G^{\varphi}$ is infinite for every automorphism $\varphi$ of $G$ (cf. \cite[Theorem 17]{bb20}, \cite[Theorem 2.4]{bb21}).
In the realm of geometric group theory, in \cite{ll00, fel01} the authors proved that the non-elementary (Gromov) hyperbolic groups have the $R_{\infty}$-property. Further, non-elementary relatively hyperbolic groups were shown to have this property by Fel'shtyn (see \cite[Theorem 3.3 (2)]{fel10}). 



Motivation of the present work comes from the results of Fel'shtyn and Gon\c{c}alves \cite{fg10}. 
The \textit{mapping class group} $\mathrm{MCG}(S)$ of  a surface $S$ is the group of isotopy classes of orientation-preserving homeomorphisms of $S$.
They proved that the mapping class group $\mathrm{MCG}(S)$ of an orientable closed surface $S$ has the $R_{\infty}$-property if and only if $S$ is not $\mathbb{S}^2$ (see \cite[Theorem 4.3]{fg10}). In the Appendix of that paper, Dahmani et al. proved that the mapping class group of the compact surfaces have the $R_{\infty}$-property except for a few low complexity cases, using geometric group theoretic methods. 
Recently, there has been a surge of activities on infinite-type surfaces and corresponding groups, namely the big mapping class groups (see \cite{bigsurvey}). 

The goal of this paper is to exhibit a similarity between the finite-type and infinite-type settings. 
Now, let $S$ be an infinite-type surface (i.e., $S$ is a connected, second countable, oriented $2$-manifold without boundary such that the fundamental group $\pi_1(S)$ is not finitely generated) and $\mathrm{MCG}(S)$ denote the big mapping class group of $S$.   
As compared to the above mentioned results (mostly algebraic in nature) our approach will significantly be more geometric. The technique from the Appendix of \cite{fg10} can be modified suitably in our set-up to show that the big mapping class group $\mathrm{MCG}(S)$ (when $S$ an infinite-type surface containing a non-displaceable subsurface) does satisfy the $R_{\infty}$-property (see \thmref{mainthmnondis}).

Using Birman exact sequence, we prove that if $S$ is an infinite-type surface with countably many isolated punctures, then the big mapping class group $\mathrm{MCG}(S)$ possesses the $R_{\infty}$-property (see \corref{maincorbes}). For details, see \secref{mcg} and \secref{results}.
As an immediate application we also characterize the $S_{\infty}$-property of the mapping class groups (see \thmref{mainthm2}). For necessary definitions and results we refer to \secref{isogred}. 

\section{Preliminaries}\label{prel}
In this section we fix some notations and terminologies, and collect some basic results which will be used throughout this paper.
\subsection{Infinite-type surfaces}
Let $S$ be a connected, orientable surface ($2$-dimensional manifold) without boundary. Then
$S$ is said to be of \textit{finite-type} if its fundamental group is finitely generated; otherwise $S$ is said to be of \textit{infinite-type}. 
Note that whenever the surface $S$ is of infinite-type the fundamental group $\pi_1(S)$ of $S$ is a free group of countable infinite rank (see, for instance, \cite[p. 142, Theorem 4.2.2]{sj93}). Finite-type surfaces (without boundary) are classified by the genus and number of punctures. We need more information to classify infinite-type surfaces, which we will discuss here. Let $S$ be an infinite-type surface.
An \textit{exiting sequence} is a collection 
$U_1 \supseteq  U_2 \supseteq \cdots $ of connected open subsets of $S$ such that
\begin{enumerate}[leftmargin=*]
	\item For any $n$, $U_n$ is not relatively compact.
	\item The boundary of $U_n$ is compact for all $n$.
	\item Every relatively compact subset of $S$ is disjoint from $U_n$ for all but
	finitely many $n$.
\end{enumerate}
Two exiting sequences are \textit{equivalent} if every element of the
first sequence is contained in some element of the second, and vice-versa.
An end of $S$ (denoted by $\mathrm{end}(S)$) is an equivalence class of exiting sequences, and we
write $\mathrm{Ends}(S)$ for the space of ends of $S$.
Note that $\mathrm{Ends}(S)$ is
totally disconnected, separable and compact space. In particular, it is a subset of a Cantor set. We refer to \cite[Section 2.1]{bigsurvey} for the topology of the end space.
An end of $S$ is called \textit{planar} if it has a representative exiting sequence
whose elements are eventually planar; otherwise it is said to be \textit{non-planar} (or \emph{accumulated by genus}).
We denote by $\mathrm{Ends}^p(S)$ and $\mathrm{Ends}^{np}(S)$ the subspaces of planar
and non-planar ends of $S$, respectively. Clearly, $\mathrm{Ends}(S) = \mathrm{Ends}^p(S)\sqcup\mathrm{Ends}^{np}(S)$. 
 Infinite-type surfaces without boundary have been classified by the following (due to Ker\'{e}k\'{j}ar\'{t}o \cite{k23} and Richards \cite{richards}): 
two connected orientable surfaces $S_1$ and $S_2$ are homeomorphic if and only if they have the same genus,
and there exists a homeomorphism $h: 
\mathrm{Ends}(S_1)\rightarrow \mathrm{Ends}(S_2)$ whose restriction to $\mathrm{Ends}^{np}(S_1)$ defines a homeomorphism
between $\mathrm{Ends}^{np}(S_1)$ and $\mathrm{Ends}^{np}(S_2)$. Note that $\text{genus}(S)<\infty$ if and only if $\mathrm{Ends}^{np}(S)=\emptyset$.
\subsection{Mapping class groups}\label{mcg}
For a surface $S$, let $\mathrm{Homeo}(S)$ denote the group of homeomorphisms on $S$. The subgroup of $\mathrm{Homeo}(S)$ consisting of homeomorphisms which are isotopic to the identity is denoted by $\mathrm{Homeo}_0(S)$ (this is also the connected component of the identity). Let $\mathrm{Homeo}^+(S)$ (resp. $\mathrm{Homeo}^-(S)$) denote the set of all orientation-preserving (resp. orientation-reversing) homeomorphisms of $S$. 
The (extended) \textit{mapping class group} of $S$, denoted by $\mathrm{MCG}^{\pm}(S)$, is the group of isotopy classes of homeomorphisms of $S$, i.e.,  $\mathrm{MCG}^{\pm}(S)=\mathrm{Homeo}(S)/\mathrm{Homeo}_0(S)$. Equivalently, it is defined as $\pi_0(\mathrm{Homeo}(S))$ with respect to the compact open topology. Now $\mathrm{MCG}^{\pm}(S)$ acts naturally by homeomorphism on the space of ends $\mathrm{Ends}(S)$, and the kernel is called pure mapping class group, denoted by $\mathrm{PMCG}^{\pm}(S)$. Both of these groups have subgroups of index $2$, denoted by $\mathrm{MCG}(S)$ and $\mathrm{PMCG}(S)$ respectively, consisting of orientation-preserving homeomorphisms. 
Notice that $\mathrm{MCG}(S)=\mathrm{Homeo}^+(S)/\mathrm{Homeo}_0(S).$ When the surface is of infinite-type, the mapping class group is homeomorphic to the Baire space, i.e., $\mathrm{MCG}(S)\cong \mathbb{R}\setminus\mathbb{Q}$. We call the mapping class group as the \textit{big mapping class group} if the surface is of infinite-type. We define a particular type of element in $\mathrm{MCG}(S)$ which will be used later. Let $A=\mathbb{S}^1\times[0,1]$ be the annulus. Define the twist map $T:A\rightarrow A$ given by $T(\theta,t)=(\theta+2\pi t, t)$ for all $(\theta,t)\in \mathbb{S}^1\times[0,1]$. Let $\alpha$ be an essential simple closed curve in $S$ (i.e., $\alpha$ is an embedding of $\mathbb{S}^1$, not homotopic to a point or a puncture). Fix a regular neighbourhood $N$ of $\alpha$ and an orientation-preserving homeomorphism $\psi: A\rightarrow N$. Now define $T_{\alpha}: S\rightarrow S$ by 
 \[T_{\alpha}(s)= \left\{
\begin{array}{ll}
\psi\circ T\circ \psi^{-1}(s) & \text{if} \; s\in N \\
s & \text{if}\; s\in S\setminus N.
\end{array}\right. \] 
The map $T_{\alpha}$ is called a \emph{Dehn twist about $\alpha$}. Note that it neither depends on $N$ nor on $\alpha$ (up to isotopy). It is an infinite order element of $\mathrm{MCG}(S)$. Now we have the following basic result. For details, we refer the reader to \cite[Chapter 3]{fm12}. 
\begin{lemma}\label{dehnt} Let $f\in\mathrm{MCG}(S)$ and $m,n\in \mathbb{N}$. Then the following hold:
	\begin{enumerate}[leftmargin=*]
	\item\label{dehnt1} $fT_{\alpha}f^{-1}=T_{f(\alpha)}$. 
	\item\label{dehnt2} $T_{\alpha}^n=T_{\beta}^m$ if and only if $\alpha=\beta$ and $m=n$.
	\end{enumerate}

\end{lemma}
\subsection{Some basic results}

In order to determine the $R_{\infty}$-property we have to understand the automorphism group of $\mathrm{MCG}(S)$ (resp. $\mathrm{MCG}^{\pm}(S)$). Following a recent work of Bavard et. al. \cite{bdr17} we will briefly describe the automorphism group of the big mapping class groups.

\noindent
\textbf{\textit{Automorphisms:}}
For a group $G$, let $\textup{Aut}(G)$ denote the group of all automorphisms of $G$, and $\mathrm{Inn}(G):=\{i_g\mid i_g(x)=gxg^{-1} \;\forall x\in G\}$, the normal subgroup of $\textup{Aut}(G)$ consisting of only the inner automorphisms of $G$. The outer automorphism group of $G$, is denoted by $\mathrm{Out}(G):=\textup{Aut}(G)/\mathrm{Inn}(G)$. A subgroup $H$ of $G$ is called \emph{characteristic} if $\varphi(H)=H$ for all $\varphi\in \textup{Aut}(G)$. 
 By virtue of \cite[Theorem 1.1]{bdr17}, for every $\varphi\in \textup{Aut}(\mathrm{MCG}^{\pm}(S))$ there exists $h\in\mathrm{Homeo}(S)$ such that $\varphi(f)=hfh^{-1}$, for all $f\in\mathrm{MCG}^{\pm}(S)$. That is every automorphism of $\mathrm{MCG}^{\pm}(S)$ is inner.  
Now let $\varphi\in \textup{Aut}(\mathrm{MCG}(S))$. Since $\mathrm{MCG}(S)$ is an index two subgroup of $\mathrm{MCG}^{\pm}(S)$ again by \cite[Theorem 1.1]{bdr17}, there exists $h\in\mathrm{Homeo}(S)$ such that $\varphi(f)=hfh^{-1}$. Now there are two possibilities for $h$: (a) $h\in\mathrm{Homeo}^+(S)$, or (b) $h\in\mathrm{Homeo}^{-}(S)$. \\
(a) If $h\in\mathrm{Homeo}^+(S)$, then $\varphi(f)=hfh^{-1}=i_{h}(f)$, i.e., $\varphi$ is an inner automorphism of $\mathrm{MCG}(S)$. (b) If $h\in\mathrm{Homeo}^{-}(S)$, then 
		$\varphi^2(f)=\varphi(hfh^{-1})=h^2fh^{-2}=i_{h^2}(f)$. Thus $\varphi^2=i_{h^2}\in \mathrm{Inn}(\mathrm{MCG}(S))$. Therefore $\overline{\varphi}$ is an element of order $1$ or $2$ in $\mathrm{Out}(\mathrm{MCG}(S))$, where $\textup{Aut}(\mathrm{MCG}(S))\rightarrow \mathrm{Out}(\mathrm{MCG}(S))$ given by $\varphi\mapsto\overline{\varphi}$. Since $\varphi$ is an arbitrary automorphism, we have  $\mathrm{Out}(\mathrm{MCG}(S))\cong \ensuremath{\mathbb{Z}}/2\ensuremath{\mathbb{Z}}\cong\langle\overline{\psi}\mid \psi: f\mapsto hfh^{-1}, h\in\mathrm{Homeo}^{-}(S)\rangle$. We summarize this as follows:
\begin{lemma}\cite[Theorem 1.1, Corollary 1.2]{bdr17}\label{out}
If $S$ is an infinite-type surface then the following hold:
\begin{enumerate}[leftmargin=*]
\item\label{out1} $\mathrm{Out}(\mathrm{MCG}^{\pm}(S))=\{e\}$.
\item\label{out2} $\mathrm{Out}(\mathrm{MCG}(S))\cong \ensuremath{\mathbb{Z}}/2\ensuremath{\mathbb{Z}}$.
\item \label{out3} If $G<\mathrm{MCG}(S)$ is a finite index subgroup then $\mathrm{Out}(G)$ is finite.
\item\label{char1} $\mathrm{MCG}(S), \mathrm{PMCG}(S)$ and $\mathrm{PMCG}^{\pm}(S)$ are characteristic subgroups in $\mathrm{MCG}^{\pm}(S)$.
\end{enumerate}
\end{lemma}  
Suppose that $1\rightarrow N \overset{i}{\rightarrow} G\overset{\pi}{\rightarrow} Q\rightarrow 1$ is an exact sequence of arbitrary groups. If $Q$ is finite and $N$ has infinitely many conjugacy classes, then we \textbf{claim} that $G$ also has infinitely many conjugacy classes.
If possible, suppose that $G$ has finitely many conjugacy classes. Since $N$ has infinitely many conjugacy classes there exist infinitely many elements $n_r\in N$ ($r\geq 0$) such that $n_r$ and $n_s$ are not conjugate in $N$ ($r\neq s$) but $n_0$ and $n_r$ are conjugate in $G$ for all $r\geq 0$. So for all $r\geq 1$ there exists $g_r\in G$ such that $n_r=g_rn_0g_r^{-1}$. 
Since $Q(\cong G/N)$ is finite there exist distinct $r,s\in \mathbb{N}$ such that $g_sg_r^{-1}\in N$. Therefore $n_s=g_sn_0g_s^{-1}=g_sg_r^{-1}n_rg_rg_s^{-1}=(g_sg_r^{-1})n_r(g_sg_r^{-1})^{-1}$. This is a contradiction to the assumption that $n_r$ and $n_s$ are not conjugate in $N$. We record this as a result below.
\begin{lemma}\label{ntog}
Let $1\rightarrow N \rightarrow G\rightarrow Q\rightarrow 1$ be an exact sequence of groups. If $Q$ is finite and $N$ has infinitely many conjugacy classes then so does $G$.
\end{lemma}
\noindent
The above lemma holds in more generality (see, \cite[Lemma 2.2(ii)]{MS}). Next, we collect the following two important results: 
\begin{lemma}\cite[Lemma 2.1]{MS}\label{qtog}
	Let $1\rightarrow N \rightarrow G\rightarrow Q\rightarrow 1$ be an exact sequence of groups. Suppose that $N$ is a characteristic subgroup of $G$ and $Q$ has the $R_{\infty}$-property, then $G$ also has the $R_{\infty}$-property.
\end{lemma}
\begin{lemma}\cite[Corollary 3.2]{FLT}\label{inner}
	Let $\varphi \in \textup{Aut}(G)$ and $i_g \in \mathrm{Inn}(G)$, then $R(\varphi\circ i_g)=R(\varphi)$. In particular, $R(i_g)=R(\mathrm{Id})$, i.e., the number of inner twisted conjugacy classes in $G$ is equal to the number of conjugacy classes in $G$.
\end{lemma}
\noindent
\textbf{\textit{Some geometric results:}}
	Let $G$ be a group acting on a metric space $(X, d)$ by isometries. Let $x\in X$ and $g\in G$. We \textbf{claim} that   $\displaystyle\lim_{n\to\infty}\frac{d(x,g^nx)}{n}$ exists. Suppose that $f:\mathbb{N} \to [0,\infty)$ is a function given by $f(n)=d(x,g^nx)$. Clearly, $f(n+m) = d(x,g^{m+n}x) \leq d(x,g^mx)+d(x,g^nx)= f(m)+f(n)$, for any $m,n\in \mathbb{N}$. Thus, $f$ is subadditive. 
	Now, let $M=\inf \frac{f(n)}{n}$. Then for any $\epsilon>0$, there exists $n>0$ such that $\frac{f(n)}{n}<M+\epsilon$. So, for every $m>n$, we have $m=kn+r$, where $k\in \mathbb{N}$ and $0\leq r<n$. So, $\frac{f(m)}{m}=\frac{f(kn+r)}{m}\leq \frac{f(kn)}{m}+\frac{f(r)}{m} \leq \frac{kn(M+\epsilon)}{m}+\frac{f(r)}{m}$.
	Therefore $\lim_{n\to\infty}\frac{d(x,g^nx)}{n}=M$. This proves the claim. For instance, see \cite[Chapter II.6, Exercise 6.6(1)]{BH}.
	
	The \emph{translation length of} $g\in G$ is given by $\tau(g):=\lim_{n\to\infty}\frac{d(x,g^nx)}{n}$. This in itself is of independent interest in geometric group theory. We have the following basic result: 
\begin{lemma}\label{translation-length}
	\begin{enumerate}[leftmargin=*]
		\item $\tau(g)$ does not depend on the choice of $x$.
		\item $\tau(g)=\tau(h^{-1}gh)$ for all $g,h \in G$.
		\item $\tau(g^k)=|k|\tau(g)$ for all $k\in \ensuremath{\mathbb{Z}}$.
	\end{enumerate}
\end{lemma}
\begin{proof}
\begin{enumerate}[leftmargin=*]
\item Let $y\in X$ be a point distinct from $x$. Then, we have\\ 
$d(x,g^nx)-d(x,y)-d(g^nx,g^ny)\leq d(y,g^ny)\leq d(y,x)+d(x,g^nx)+d(g^nx,g^ny)$.\\
This implies $\frac{d(x,g^nx)}{n}-2\frac{d(x,y)}{n}\leq \frac{d(y,g^ny)}{n}\leq\frac{d(x,g^nx)}{n}+2\frac{d(x,y)}{n}$.\\ 
Thus, we have $\lim_{n\to\infty}\frac{d(x,g^nx)}{n} =\lim_{n\to\infty}\frac{d(y,g^ny)}{n}$.
\item For any $g, h\in G$, we have $$\tau(h^{-1}gh)= \lim_{n\to\infty}\frac{d(x,(h^{-1}gh)^nx)}{n}=\lim_{n\to\infty}\frac{d(x,h^{-1}g^nhx)}{n}=\lim_{n\to\infty} \frac{d(hx,g^n(hx))}{n}.$$ 
Then by $(1)$, $\tau(h^{-1}gh)=\tau(g)$.
\item Assume that $k>0$, then $$\tau(g^k)=\lim_{n\to\infty}\frac{d(x,(g^k)^nx)}{n}=\lim_{n\to\infty}\frac{kd(x,g^{nk}x)}{nk}=k\lim_{nk\to\infty}\frac{d(x,g^{nk}x)}{nk}=k\tau(g).$$ Similarly one can check this for $k<0$. Hence we have the result.	\end{enumerate}
\end{proof}
\noindent
The above result immediately provides a sufficient condition for a group to have infinitely many conjugacy classes (cf. \cite[Corollary 8.2.G]{gromov}). An isometry $g$ is called \emph{loxodromic} if $\tau(g)>0$. 
\begin{corollary}\label{infinitecc}
	If a metric space $(X,d)$ admits an isometric $G$-action such that $\tau(g)>0$, then the number of conjugacy classes of $G$ is infinite.
\end{corollary}
\begin{proof}
First, note that the order of $g$ is infinite because $\tau(g)>0$. Now look at the infinite sequence $\{g^i\}_{i\in \mathbb{N}}$ of distinct elements of $G$. By \lemref{translation-length} $g^i$ is not conjugate to $g^j$ (for $i\neq j$). Thus, $G$ has infinitely many conjugacy classes.
\end{proof}
\noindent
In order to describe the next set of results we fix some notations.
Let $\varphi\in\textup{Aut}(G)$ and $\ensuremath{\mathbb{Z}}=\langle t\rangle$. Define the following extension of $G$ containing $G$ as a normal subgroup:
$$G_{\varphi}:=\langle G,t\mid tgt^{-1}=\varphi(g)\;  \forall g\rangle=G\rtimes_{\varphi} \ensuremath{\mathbb{Z}}=\bigsqcup_{n\in\ensuremath{\mathbb{Z}}} G\cdot t^n.$$ 
Now that we have the notations laid out, we state the following: 
\begin{lemma}\cite[Lemma 2.1]{fel10}\label{keylemma}
	Two elements $x$ and $y$ of $G$ are $\varphi$-twisted conjugate if and only if $xt$ and $yt$ of $G_{\varphi}$ are conjugate in $G_{\varphi}$. 
\end{lemma}
\begin{proof}
	If $x$ and $y$ are $\varphi$-twisted conjugate then $ y=gx\varphi(g^{-1})=gxtg^{-1}t^{-1}$ for some $g\in G$. Then $yt=g(xt)g^{-1}$.
	
	Conversely, suppose that $xt$ and $yt$ are conjugate in $G_{\varphi}$. Then there exists an element $gt^n$ in $G_{\varphi}$ such that   $yt=(gt^n)(xt)(gt^n)^{-1}=g(t^nxt^{-n})tg^{-1}=g\varphi^n(x)tg^{-1}.$ Thus $y=g\varphi^n(x)(tg^{-1}t^{-1})=g\varphi^n(x)\varphi(g^{-1})$, i.e., $y\sim_{\varphi} \varphi^n(x)$. Also, note that $x$ and $\varphi(x)$ are always $\varphi$-twisted conjugate. Thus, $x\sim_{\varphi}y$.
\end{proof}
\noindent
This gives an immediate corollary which will be very useful later.
\begin{corollary}\label{bijection}
	The map $x\mapsto xt$ gives a bijection between the set of $\varphi$-twisted conjugacy classes of $G$ and the set of conjugacy classes of $G_{\varphi}$ contained in the coset $G\cdot t$. That is  $R(\varphi)$ is the number of conjugacy classes  of $G_{\varphi}$ in the coset $G\cdot t$ of $G$.
\end{corollary}
In a more general set-up we have the following result due to T. Delzant (cf. \cite[Lemma 6.3]{fg10},\cite[Lemma 3.4]{ll00}). Let $G$ be a group acting on a hyperbolic metric space $X$ by isometries. The limit set $\Lambda(G)$ of $G$ is defined as $\Lambda(G):=\overline{Gx}\cap \partial X$ (for some $x\in X$), where $\partial X$ denotes the Gromov boundary of $X$. The action is called \emph{non-elementary} if $|\Lambda(G)|=\infty$. Note that if the action is non-elementary then there exists a loxodromic isometry. For details, see \cite[Section 8.2.D]{gromov}.
\begin{lemma}\cite[Lemma 3.4]{ll00}\label{delzant}
	Let $N$ be a normal subgroup of $G$ such that $G/N$ is abelian. If $G$ acts on a hyperbolic space non-elementarily, then any coset of $N$ contains infinitely many conjugacy classes of $G$.
\end{lemma}
Now we recall Alexander method for infinite-type surfaces.
\begin{lemma}\cite[Corollary 1.2]{hmv19}\label{alexander}
	Suppose that $S$ is an infinite-type surface. If $f\in \mathrm{MCG}^{\pm}(S)$ fixes each essential simple closed curve of $S$ then $f=\mathrm{Id}$.
\end{lemma}

\begin{lemma}\label{center}
	Let $S$ be a connected infinite-type surface without boundary then the centers $Z(\mathrm{MCG}^{\pm}(S))=\{\mathrm{Id}\}=Z(\mathrm{MCG}(S))$.
\end{lemma}
\begin{proof}
	Lanier and Loving \cite[Proposition 2]{ll20} proved that $\mathrm{MCG}(S)$ is centerless. 
	In fact by the same line of argument we prove that $Z(\mathrm{MCG}^{\pm}(S))=\{\mathrm{Id}\}$, using the Alexander method.
	If possible suppose that $\mathrm{Id}\neq f\in Z(\mathrm{MCG}^{\pm}(S))$. Then by \lemref{alexander} there exists a curve $\gamma$ such that $f(\gamma)\neq \gamma$. Now let $T_{\gamma}\in \mathrm{MCG}^{\pm}(S)$ be a Dehn twist about $\gamma$. Then $fT_{\gamma}f^{-1}=T_{\gamma}$ as $f$ is central. Therefore $T_{f(\gamma)}=T_{\gamma}$, which in turn implies that $f(\gamma)=\gamma$ (by \lemref{dehnt}). We get a contradiction to the fact that $f(\gamma)\neq \gamma$. Therefore the (extended) big mapping class group $\mathrm{MCG}^{\pm}(S)$ is also  centerless.
\end{proof}
\noindent
We end this section with the following result. 
Let $C$ be a Cantor space.
By a result of Rubin \cite[p. 492]{rubin}, every automorphism of $\mathrm{Homeo}(C)$ is inner\footnote{This result has been brought to our notice by Yves de Cornulier}. Also, note that 
$\mathrm{Homeo}(C)$ has infinitely many (more precisely uncountably many) conjugacy classes (see \cite[Section 5, Theorem IV]{and58}). Therefore by \lemref{inner} $\mathrm{Homeo}(C)$ has the $R_{\infty}$-property. We record this discussion as follows:
\begin{lemma}\label{cantorr}
	Let $X$ be a topological space homeomorphic to the Cantor space, then $\mathrm{Homeo}(X)$ possesses the $R_{\infty}$-property.
\end{lemma}





\section{Main results}\label{results}
\noindent
We begin with the following basic result.
\begin{theorem}\label{mainthm1}
Let $S$ be a connected orientable infinite-type surface without boundary. Then the $($extended$)$ big mapping class group $\mathrm{MCG}^{\pm}(S)$ has the $R_{\infty}$-property.
\end{theorem}
\begin{proof}
Let $T_{\alpha}$ be a Dehn twist (with respect to an essential simple closed curve $\alpha$). Then $T_{\alpha}\in \mathrm{MCG}(S)$ is an infinite order element. Consider the infinite sequence $\{T_{\alpha}^n\mid n\in \mathbb{N}\}$ of distinct elements of $\mathrm{MCG}(S)$. For distinct $m,n\in \mathbb{N}$ if $T_{\alpha}^n$ is conjugate to $T_{\alpha}^m$ in $\mathrm{MCG}(S)$, then $fT_{\alpha}^nf^{-1}=T_{\alpha}^m$ for some $f\in \mathrm{MCG}(S)$. This implies that $T_{f(\alpha)}^n=T_{\alpha}^m$ (by \lemref{dehnt}\eqref{dehnt1}). Therefore, by \lemref{dehnt}\eqref{dehnt2} we have $f(\alpha)=\alpha$ and $m=n$, which is a contradiction. Hence $\mathrm{MCG}(S)$ has infinitely many conjugacy classes (of representatives $T_{\alpha}^n$ for each $n\in \mathbb{N}$). 
Now, look at the following short exact sequence of groups:
\[1\rightarrow\mathrm{MCG}(S)\rightarrow\mathrm{MCG}^{\pm}(S)\rightarrow\ensuremath{\mathbb{Z}}/2\ensuremath{\mathbb{Z}}\rightarrow 1.\]
Then by \lemref{ntog} $\mathrm{MCG}^{\pm}(S)$ has infinitely many conjugacy classes. Therefore by \lemref{inner} and \lemref{out}\eqref{out1} it follows that the (extended) big mapping class group $\mathrm{MCG}^{\pm}(S)$ possesses the $R_{\infty}$-property.
\end{proof}
\noindent
There is an important well-studied hyperbolic space (namely the curve graph $\mathcal{C}(S)$) on which the mapping class group of a finite-type surface $S$ acts non-elementarily. But for the infinite-type surfaces there is no such one space on which big mapping class group acts non-elementarily. So we shall deal with this case-by-case.
For the rest of the paper, we will only consider the big mapping class group $\mathrm{MCG}(S)$.
\subsection{Surfaces with non-displaceable subsurfaces}
A finite-type connected subsurface $\Sigma\subset S$ is called \textit{non-displaceable} if $f(\Sigma)\cap \Sigma\neq \emptyset$ for all $ f\in \mathrm{Homeo}(S)$.
In \cite[Theorem 3.13]{rafi}, the authors also gave an algebraic characterisation which says the following: $\Sigma\subset S$ is non-displaceable if and only if $\mathrm{MCG}(S)$ contains a non-trivial normal free subgroup. It follows from the classification of surfaces that there are uncountably many such infinite-type surfaces.
\begin{example}\label{ex}
	\begin{enumerate}[leftmargin=*]
		\item\label{finitegenus}
		Let $0<\text{Genus}(S)<\infty$. By the classification of surfaces, we have $S\cong \Sigma_g-E$, where $E\simeq\mathrm{Ends}(S)$ is a totally disconnected closed subset of $\Sigma_g$ (see \cite{richards}). 
		Then $\Sigma_g$ is a non-displaceable subsurface of $S$ as $\Sigma_g$ contains non-separating curves but $S-\Sigma_g$ does not.
		\item Let $S:=\Omega_n\;(n\geq 3)$ be an infinite genus surface with exactly $n$ non-planar ends then $S$ contains a non-displaceable subsurface.
		\item If $S$ is a zero genus surface such that $\mathrm{Ends}(S)$ is countable and $|\mathrm{Ends}(S)|\geq 3$, then $S$ contains a non-displaceable subsurface.
		\item The Blooming Cantor tree surface also contains a non-displaceable subsurface.
	\end{enumerate} 
\end{example}
For more examples, the reader is referred to \cite{mr20}. In order to state and prove one of our main theorems, we need one more piece of information. There is a celebrated construction of quasi-trees of metric spaces (popularly known as \emph{BBF family for a group $G$}) developed by Bestvina, Bromberg and Fujiwara. We will not be describing this here but we refer the interested readers to \cite{BBF}. However, using this machinery, in \cite{rafi} the authors proved the following important result which will be sufficient for our purpose. 
\begin{lemma}\cite[Theorem 2.9]{rafi}\label{hqr}
Let $S$ be a connected orientable infinite-type surface which contains a connected non-displaceable subsurface $\Sigma$ of finite-type. Then there exists an unbounded hyperbolic space $X$  equipped with a $($continuous$)$ non-elementary isometric action of $\mathrm{MCG}(S)$ $($resp. $\mathrm{MCG}^{\pm}(S))$.
\end{lemma}
\begin{theorem}\label{mainthmnondis}
	Let $S$ be a connected orientable infinite-type surface without boundary which contains a non-displaceable subsurface then $\mathrm{MCG}(S)$ has the $R_{\infty}$-property.
\end{theorem}
\begin{proof}
 In view of \lemref{hqr}, the action of $\mathrm{MCG}(S)$ (resp. $\mathrm{MCG}^{\pm}(S)$) on $X$ is non-elementary. Then there is an element $g$ of $\mathrm{MCG}(S)$ such that $\tau(g)>0$ (i.e., $g$ is a loxodromic isometry with respect to this action). By \corref{infinitecc}, $\mathrm{MCG}(S)$ has infinitely many conjugacy classes (of loxodromic elements)\footnote{$\mathrm{MCG}(S)$ has infinitely many conjugacy classes by powers of a Dehn twist also as in \thmref{mainthm1}.}. Now let $\varphi\in \textup{Aut}(\mathrm{MCG}(S))$ which is not inner. Then by \lemref{out}\eqref{out2}, we have 
	$\mathrm{MCG}(S)_{\varphi}\cong\mathrm{MCG}(S)\rtimes_{\varphi}\ensuremath{\mathbb{Z}}/2\ensuremath{\mathbb{Z}}\cong\mathrm{MCG}^{\pm}(S)$, i.e., we have the following short exact sequence of groups
	$$1\rightarrow\mathrm{MCG}(S)\rightarrow \mathrm{MCG}(S)_{\varphi}\rightarrow \ensuremath{\mathbb{Z}}/2\ensuremath{\mathbb{Z}}\rightarrow 1,$$
	where $\ensuremath{\mathbb{Z}}/2\ensuremath{\mathbb{Z}}=\langle t\in \mathrm{Homeo}^{-}(S)\mid t^2=_{\mathrm{MCG}(S)}\mathrm{Id}\rangle$. 
	Therefore by \lemref{delzant}, $\mathrm{MCG}(S)$ contains infinitely many conjugacy classes of $\mathrm{MCG}^{\pm}(S)$.
	Hence by \corref{bijection}, $R(\varphi)=\infty$, which in turn implies that $\mathrm{MCG}(S)$ has the $R_{\infty}$-property.
\end{proof}
\begin{corollary}
With notations as in \thmref{mainthmnondis}, let $G<\mathrm{MCG}(S)$ be a finite index subgroup. Then $G$ has the $R_{\infty}$-property. 
\end{corollary}
\begin{proof}
Since $G$ is a finite index subgroup of $\mathrm{MCG}(S)$, then $G$ acts non-elementarily on the hyperbolic space $X$ (\lemref{hqr}). Thus, there exists a loxodromic element $g\in G$ such that $\tau(g)>0$. By \corref{infinitecc} $G$ has infinitely many conjugacy classes. Now by \lemref{out}\eqref{out3} $\mathrm{Out}(G)$ is finite. Hence the proof follows by virtue of \corref{bijection}.
\end{proof}
\subsection{Surfaces without non-displaceable subsurfaces}
From now on assume that an infinite-type surface $S$ does not contain a non-displaceable subsurface. Thus, in view of Example \ref{ex}\eqref{finitegenus}, $S$ has either zero genus or infinite genus. Also, by \cite[Theorem 6.1]{apv} either the end space $\mathrm{Ends}(S)$ of $S$ is self-similar or it is doubly pointed. For example, the Loch Ness Monster surface is self-similar and the Jacob ladder surface is doubly pointed. In order to deal with this we discuss the following general construction originally due to Birman.
\subsection{Birman exact sequence:}
Let $S$ be a connected orientable surface (both finite-type and infinite-type) without boundary. Fix $x\in S$, then  $f\in \mathrm{Homeo}^+(S)$ evaluated at $x$ gives rise to the following \emph{fiber bundle}
$$\mathrm{Homeo}^+(S,x)\longrightarrow\mathrm{Homeo}^+(S)\overset{p}{\longrightarrow}S,$$ 
where $p(f)=f(x)$ and $\mathrm{Homeo}^+(S,x)$ is the set of all orientation preserving homeomorphisms of $S$ which fix $x$. Then it induces the following homotopy long exact sequence 
\begin{center}
	
	\begin{tikzpicture}[descr/.style={fill=white,inner sep=1.5pt}]
	\matrix (m)[matrix of math nodes,row sep=1em,column sep=2.4em,text height=1.5ex, text depth=0.25ex]
	{ \cdots & \pi_1(\mathrm{Homeo}^+(S,x)) & \pi_1(\mathrm{Homeo}^+(S)) & \pi_1(S,x) \\
		& \pi_0(\mathrm{Homeo}^+(S,x)) & \pi_0(\mathrm{Homeo}^+(S)) & \pi_0(S) & \cdots. \\};
	\path[overlay,->, font=\scriptsize,>=latex]
	(m-1-1) edge (m-1-2)
	(m-1-2) edge (m-1-3)
	(m-1-3) edge (m-1-4)
	(m-1-4) edge[out=355,in=175] node[descr,yshift=0.3ex] {} (m-2-2)
	(m-2-2) edge (m-2-3)
	(m-2-3) edge (m-2-4)
	(m-2-4) edge (m-2-5);
	\end{tikzpicture}
\end{center}

Note that the group $\pi_1(\mathrm{Homeo}^+(S))$ is trivial (if $S$ is of infinite-type see \cite[Theorem 1.1]{y00} and if $S$ is finite-type see \cite[Theorem 1.14]{fm12}), and $\pi_0(S)=\{e\}$ as $S$ is connected. Therefore we have
\begin{align}\label{birman}
1\longrightarrow\pi_1(S,x)\overset{\mathcal{P}}{\longrightarrow}\mathrm{MCG}(S\setminus x)\overset{\mathcal{F}}{\longrightarrow}\mathrm{MCG}(S)\longrightarrow 1,
\end{align}
where the first map $\mathcal{P}$ is called a point pushing map and the last map $\mathcal{F}$ is the  forgetful map. Note that $\pi_1(S,x)\cong \pi_0(\mathrm{Homeo}^+(S,x)\cap\mathrm{Homeo}_0(S))$ (see \cite[Section 1.2.3]{mj11}). 
We refer the reader to \cite[p. 100, Section 4.2.3]{fm12} for more details of the above discussion. It's called  the \emph{Birman exact sequence}.  

\begin{lemma}\label{char-lemma}
	The fundamental group $\pi_1(S,x)$ is a characteristic subgroup of $\mathrm{MCG}(S\setminus x)$.
\end{lemma}
\begin{proof}
	As Equation \eqref{birman} is a short exact sequence, we have $\mathrm{ker}(\mathcal{F})=\mathrm{Im}(\mathcal{P}) =\pi_1(S,x)$. First, note that $\pi_1(S,x)$ is a normal subgroup, i.e., it is invariant under every inner automorphism of $\mathrm{MCG}(S\setminus x)$.   Now let $\varphi\in\textup{Aut}(\mathrm{MCG}(S\setminus x))$ which is not inner. By the discussion of \lemref{out} there exists an $h\in \mathrm{Homeo}^{-}(S\setminus x)$ such that $\varphi(f)=hfh^{-1}$ for all $f\in \mathrm{MCG}(S\setminus x)$. 
	Let $f\in \mathrm{MCG}(S\setminus x)$ such that $\mathcal{F}(f)=\mathrm{Id}$. 
	
	\textbf{Claim:} $\varphi(f)\in \mathrm{ker}(\mathcal{F})$.
	
	Let $\psi\in \mathrm{Homeo}^+(S\setminus x)$ be an element representing $f$, i.e., $[\psi]=f$. Suppose  $\overline{\psi}$ is the image of $\psi$ in $\mathrm{Homeo}^+(S)$. Then $\overline{\psi}$ is isotopic to the identity homeomorphism of $S$, i.e., $[\overline{\psi}]=\mathrm{Id}$ since $f\in \mathrm{ker}(\mathcal{F})$.
	Now, $h\circ\psi\circ h^{-1}$ is an orientation preserving homeomorphism of $S\setminus x$ representing $h\circ f\circ h^{-1}$ (as both $h$ and $\psi$ fix $x$). Therefore $\overline{h\psi h^{-1}}$ is isotopic to the identity as $\overline{\psi}$ is so, i.e., $[\overline{h\psi h^{-1}}]=\mathrm{Id}$. Thus $\varphi(f)\in \mathrm{ker}(\mathcal{F})$.
\end{proof}
\begin{corollary}\label{stosx}
	If $\mathrm{MCG}(S)$ has the $R_{\infty}$-property then $\mathrm{MCG}(S\setminus x)$ also has the $R_{\infty}$-property.  
\end{corollary}
\begin{proof}
	This follows from Equation \eqref{birman}, \lemref{qtog}, and \lemref{char-lemma}.
\end{proof}
\begin{corollary}\label{maincorbes}
	Suppose that $S$ has countably many isolated punctures, and let $S^*$ be $S$ with countably many marked points. Then $\mathrm{MCG}(S^*)$ has the $R_{\infty}$-property, provided $\mathrm{MCG}(S)$ does.
\end{corollary}
\begin{proof}
	This follows repeated application of \corref{stosx} and  Equation \eqref{birman}.  
\end{proof}
\begin{corollary}
Suppose that $\Sigma_g^n$ $(g\geq 1)$ is a finite-type surface with $g$ genus and $n$ punctures, then $\mathrm{MCG}(\Sigma_g^n)$ has the $R_{\infty}$-property.
\end{corollary}
\noindent
Now, let $C$ denote the Cantor set.
\begin{lemma}\label{s2-c}
	Then $\mathrm{MCG}(\mathbb{S}^2\setminus C)$ has the $R_{\infty}$-property.
\end{lemma}
\begin{proof}
	Observe that $\mathrm{Ends}(\mathbb{S}^2\setminus C)$ is homeomorphic to $C$. The mapping class group $\mathrm{MCG}(\mathbb{S}^2\setminus C)$ acts on its space of ends $C$ by homeomorphism. Which in turn induces the following short exact sequence of groups 
	\[1\rightarrow\mathrm{PMCG}(\mathbb{S}^2\setminus C)\rightarrow\mathrm{MCG}(\mathbb{S}^2\setminus C)\rightarrow\mathrm{Homeo}(C)\longrightarrow 1.\]
Now, in view of \lemref{cantorr}, $\mathrm{Homeo}(C)$ possesses the $R_{\infty}$-property. Then by \lemref{qtog}, $\mathrm{MCG}(\mathbb{S}^2\setminus C)$ has the $R_{\infty}$-property as $\mathrm{PMCG}(\mathbb{S}^2\setminus C)$ is a  characteristic subgroup of $\mathrm{MCG}(\mathbb{S}^2\setminus C)$. 
\end{proof}
\begin{corollary}\label{r2-c}
	The mapping class group $\mathrm{MCG}(\mathbb R^2\setminus C)$ has the $R_{\infty}$-property.
\end{corollary}
\begin{proof}
	Look at the following Birman exact sequence of groups (cf. Equation \eqref{birman}) 
	\[1\rightarrow\pi_1(\mathbb{S}^2\setminus C, x)\overset{\mathcal{P}}{\rightarrow}\mathrm{MCG}(\mathbb{R}^2\setminus C)\overset{\mathcal{F}}{\rightarrow}\mathrm{MCG}(\mathbb{S}^2\setminus C)\rightarrow 1.\] 
 Now note that $\pi_1(\mathbb{S}^2\setminus C, x)$ is a characteristic subgroup $\mathrm{MCG}(\mathbb R^2\setminus C)$ (by \lemref{char-lemma}). By \lemref{s2-c}, $\mathrm{MCG}(\mathbb{S}^2\setminus C)$ has the $R_{\infty}$-property. Hence the result follows from \lemref{qtog}.  
\end{proof}
\begin{remark}
	By a result of Bavard $\mathrm{MCG}(\mathbb R^2\setminus C)$ acts non-elementarily on a hyperbolic space called ray graph. Thus, the $R_{\infty}$-property of $\mathrm{MCG}(\mathbb{R}^2\setminus C)$ also follows by \corref{bijection} and \lemref{delzant}.
\end{remark}
\noindent
We end this section with the following important class of groups which has the $R_{\infty}$-property.
\begin{theorem}
Let $S$ be an infinite-type surface such that the end space $\mathrm{Ends}(S)$ is a Cantor set $C$ with the following: either (a) $\mathrm{genus}(S)=0$ or (b) $\mathrm{Ends}(S)=\mathrm{Ends}^{np}(S)$.
Then $\mathrm{MCG}(S)$ possesses the $R_{\infty}$-property.
\end{theorem}
\begin{proof}
If $\mathrm{genus}(S)=0$, then $\mathrm{Ends}^{np}(S)=\emptyset$. So in both the cases we get the following short exact sequence of groups (cf. \cite[Section 2.4.1]{bigsurvey})
\[1\rightarrow\mathrm{PMCG}(S)\rightarrow\mathrm{MCG}(S)\rightarrow\mathrm{Homeo}(C)\longrightarrow 1.\] By \lemref{cantorr}, $\mathrm{Homeo}(C)$ satisfies the $R_{\infty}$-property. Hence by \lemref{qtog}, $\mathrm{MCG}(S)$ satisfies the $R_{\infty}$-property since $\mathrm{PMCG}(S)$ is a characteristic subgroup.
\end{proof}
\begin{remark}
For example, the Blooming Cantor tree surface ($\mathrm{genus}(S)=\infty$ and $\mathrm{Ends}(S)=\mathrm{Ends}^{np}(S)=C$) and the Cantor tree surface ($\simeq\mathbb{S}^2\setminus C$ and $\mathrm{genus}(S)=0$) satisfy the hypothesis of the previous theorem. See \cite{bigsurvey}, for details.
\end{remark}
\section{Isogredience classes}\label{isogred}
Most of what is contained in this section is known and can mainly be found in \cite{ft15}, \cite{bdr21} and the references therein. Still, it seems worthwhile to combine several results in a fashion that suits our purpose and make the paper more self-contained. 
Suppose $\Phi\in \mathrm{Out}(G):=\mathrm{Inn}(G)\backslash\textup{Aut}(G)$. Two elements $\alpha, \beta\in \Phi$ are said to be \emph{isogredient} (or similar) if $\beta=i_h\circ \alpha \circ i_{h^{-1}}$ for some $h\in G$, where $i_h(g)=hgh^{-1}$ for all $g\in G$. Observe that  this is an equivalence relation on $\Phi$. Fix a representative $\gamma\in \Phi$. Then $\alpha=i_a\circ \gamma$ and $\beta=i_b\circ \gamma$ for some $a,b\in G$. Therefore 
\begin{align*}
i_b\circ \gamma&=\beta=i_h\circ \alpha \circ i_{h^{-1}}
=i_h\circ i_a\circ \gamma \circ i_{h^{-1}}\\
i_b&=i_h\circ i_a\circ (\gamma \circ i_{h^{-1}}\circ\gamma^{-1})\\
i_b&=i_h\circ i_a\circ i_{\gamma(h^{-1})}\\
i_b&=i_{ha\gamma(h^{-1})}.
\end{align*}
Thus, $\alpha$ and $\beta$ are isogredient if and only if $b=ha\gamma(h^{-1})c$ for some $c\in Z(G)$, the center of $G$.
Let $S(\Phi)$ denote the number of isogredience classes of $\Phi$. If $\Phi=\overline{\textup{Id}}$, then $S(\overline{\textup{Id}})$ is the number of ordinary conjugacy classes of $G/Z(G)$. A group $G$ has the \emph{$S_{\infty}$-property} if $S(\Phi)=\infty$ for all $\Phi\in \mathrm{Out}(G)$. Observe that if a group $G$ has the $S_{\infty}$-property then $G/Z(G)$ possesses the $R_{\infty}$-property, which in turn implies that $G$ satisfies the $R_{\infty}$-property (by \lemref{qtog}). We summarize this discussion as follows:
\begin{lemma}\cite[cf. Theorem 3.4]{ft15}\label{rvss}
	\begin{enumerate}[leftmargin=*]
	\item\label{1} If $G$ satisfies the $S_{\infty}$-property then $G$ satisfies the $R_{\infty}$-property.
	\item\label{2}  If $G$ is centerless and $G$ has the $R_{\infty}$-property then $G$ has the $S_{\infty}$-property.
	\end{enumerate}
	
\end{lemma} 
Levitt and Lustig \cite[Theorem 3.5]{ll00} proved that any non-elementary hyperbolic group has the $S_{\infty}$-property. Torsion-free non-elementary hyperbolic group is centerless, therefore it satisfies the $S_{\infty}$-property if and only if it satisfies the $R_{\infty}$-property. 
\begin{remark}
	Even before, Cohen and Lustig \cite[Proposition 5.4]{cl89} studied the $S_{\infty}$-property for free group $\mathbb{F}_n$ although they did not use the terminology $S_{\infty}$. Best to our knowledge, the term `group with the $S_{\infty}$-property' was introduced by Fel'shtyn and Troitsky in \cite{ft15}, especially see Lemma 3.3 and Theorem 3.4 for interplay between the $R_{\infty}$-property and $S_{\infty}$-property.
\end{remark}
\noindent
Now we are in a position to prove our main result of this section.
\begin{theorem}\label{mainthm2}
	Let $S$ be a connected orientable infinite-type surface without boundary. Then we have the following:
	\begin{enumerate}[leftmargin=*]
	\item the $($extended$)$ big mapping class group $\mathrm{MCG}^{\pm}(S)$ possesses the $S_{\infty}$-property.
	\item If $S$ contains a connected non-displaceable subsurface of finite-type then $\mathrm{MCG}(S)$ has the $S_{\infty}$-property.
	\item $\mathrm{MCG}(S)$ has the $R_{\infty}$-property if and only if $\mathrm{MCG}(S)$ has the $S_{\infty}$-property.
	\end{enumerate}
\end{theorem}
\begin{proof}
	\begin{enumerate}[leftmargin=*]
	\item In view of \thmref{mainthm1}, $\mathrm{MCG}^{\pm}(S)$ has the $R_{\infty}$-property. Now by virtue of \lemref{center} we have $Z(\mathrm{MCG}^{\pm}(S))=\{\mathrm{Id}\}$. Therefore by \lemref{rvss}\eqref{2} $\mathrm{MCG}^{\pm}(S)$ possesses the $S_{\infty}$-property.
	\item By \thmref{mainthmnondis} $\mathrm{MCG}(S)$ satisfies the $R_{\infty}$-property. In view of  \lemref{center} the center is trivial, i.e.,  $Z(\mathrm{MCG}(S))=\{\mathrm{Id}\}$. Therefore by \lemref{rvss}\eqref{2}, $\mathrm{MCG}(S)$ has the $S_{\infty}$-property.
	\item This follows from \lemref{rvss} and \lemref{center}. 
	\end{enumerate}
\end{proof}
\begin{remark}
If $S$ is a finite-type surface without boundary (except few cases) then we know that the center $Z(\mathrm{MCG}(S))=\{\mathrm{Id}\}$ (see \cite[Theorem 3.10]{fm12}). Now $\mathrm{MCG}(S)$ satisfies the $R_{\infty}$-property by \cite[cf. Theorem 6.4, Theorem 4.3]{fg10}. So, again  in view of \lemref{rvss}, $\mathrm{MCG}(S)$ has the $S_{\infty}$-property.
\end{remark}
\noindent
We would like to end this paper with the following problem.

\medskip
\noindent
\textbf{Problem:} Suppose that $S$ is either the Loch Ness Monster surface or the Jacob Ladder surface. Does $\mathrm{MCG}(S)$ satisfy the $R_{\infty}$-property? 

\medskip
\noindent
\textbf{Acknowledgement:} The authors would like to thank Pranab Sardar for bringing to our notice about infinite-type surfaces and his encouragement throughout this work. We thank Anirban Bose for some useful discussions. 


