

\section{Supplemental material}
\subsection{A. Proof of Theorem~\ref{main_thm}}

\begin{lemma}\label{monomial}
    For $g \in \mathcal{M}_m$,  
    \begin{equation}
        [B_n(g)]_{\vb*{n'}\vb*{n}}=\delta_{\sigma_g(\vb*{n'}),\vb*{n}}  \prod_{i=0}^{m-1} [D(g)]_{ii}^{n_i}.
        \label{monomial_formula}
    \end{equation}
\end{lemma}
\begin{proof}
Because $B_n$ is a group homomorphism, $B_n(g)= B_n(P_{\sigma_g})B_n(D(g))$.
From the definition, 
\begin{equation}\label{permutation_formula}
    B(P_{\sigma_g})\ket*{\vb*{n}}=\prod_{i}\frac{1}{\sqrt{n_i!}}\qty\Big(\sum_{j}(a^\dag_j [P_{\sigma_g}]_{ji}))^{n_i}\ket*{\vb*{0}} 
    =\prod_{i}\frac{1}{\sqrt{n_i!}}(a^\dag_{\sigma_g(i)})^{n_i}\ket*{\vb*{0}} = \ket*{{\sigma_g}^{-1}(\vb*{n})}
\end{equation}
and
\begin{equation}\label{diagonal_formula}
    B(D(g))\ket*{\vb*{n}}=\prod_{i}\frac{1}{\sqrt{n_i!}}\qty\Big(\sum_{j}(a^\dag_j [D(g)]_{ji}))^{n_i}\ket*{\vb*{0}} 
    =\prod_{i}[D(g)]_{ii}^{n_i}\ket*{\vb*{n}}.
\end{equation}
Therefore, we have Eq.~\eqref{monomial_formula}.
\end{proof}

\begin{lemma}\label{projector}
Let 
\begin{equation}
    \rho^G_\lambda = \frac{1}{\abs{G}}\sum_{g\in G} \lambda(g)^{-1} B(g)
\end{equation}
for an Abelian group $G$. Then, $\rho^G_\lambda$ is the projector onto $V^G_\lambda$.
\end{lemma}
\begin{proof}
For any state $\ket*{\Psi} \in V^G_\lambda$, $\rho^G_\lambda \ket*{\Psi}=\ket*{\Psi}$ holds.
It also holds that 
\begin{align}
    B(g) \rho^G_\lambda &= \frac{1}{\abs{G}}\sum_{h\in G} \lambda(h)^{-1} B(gh) 
    =\frac{1}{\abs{G}}\sum_{h\in G} \lambda(g^{-1}h)^{-1} B(h) =
    \lambda(g)\rho^G_\lambda 
\end{align}
for any $g \in G$.
Therefore, $\rho^G_\lambda \ket*{\Psi} \in  V^G_\lambda$ for any state $\ket*{\Psi}$, and $(\rho^G_\lambda)^2 =\rho^G_\lambda$ .
\end{proof}
Lemma~\ref{projector} can be regarded as a special case of the projection formula in representation theory~\cite{Fulton1991-gz}, where the considered group is Abelian.
\begin{proof}[Proof of Theorem~\ref{main_thm}]
(i) We have
\begin{align}
    \norm{\rho_\lambda^G \ket*{\Psi}}^2 &= \sum_{\vb*{n},\vb*{n'}}  c_{\vb*{n'}}^*  c_{\vb*{n}} \bra{\vb*{n'}} \rho_\lambda^G \ket*{\vb*{n}}
    = \sum_{\vb*{n},\vb*{n'}}  c_{\vb*{n'}}^*  c_{\vb*{n}} \frac{1}{\abs{G}}\sum_{g\in G}\lambda(g)^{-1} \bra{\vb*{n'}} B(g) \ket*{\vb*{n}} \notag \\
    &= \frac{1}{\abs{G}}\sum_{g\in G} \lambda(g)^{-1}\sum_{\vb*{n}}c_{\sigma^{-1}_g(\vb*{n})}^* c_{\vb*{n}}\prod_{i=0}^{m-1} D(g)_{i,i}^{n_i},
\end{align}
where the second and third equalities hold from Lemmas~\ref{projector} and~\ref{monomial}, respectively.

(ii) If $\ket*{\Psi_\text{in}}\in V_\lambda^G$ holds, $\ket*{\Psi_\text{out}}=B(U)\ket*{\Psi_\text{in}}$ satisfies 
\begin{equation}
    B(UgU^\dag)B(U)\ket*{\Psi_\text{in}}=B(U)B(g)\ket*{\Psi_\text{in}}=\lambda(g)B(U)\ket*{\Psi_\text{in}}
\end{equation}
for any $g\in G$. 
Thus, letting $G'=UGU^\dag$ and $\lambda'(UgU^\dag)=\lambda(g)$ for $g \in G$, it holds that  $\ket*{\Psi_\text{out}} \in V^{G'}_{\lambda'}$. 
The converse also holds.

(iii) $\bra{\vb*{n}}\ket*{\Psi}=\bra{\vb*{n}}\rho_\lambda^G \ket*{\Psi}=0$ if $\rho_\lambda^G\ket*{\vb*{n}}=0$. 
For any $g\in G$, 
\begin{equation}
    \lambda(g)\bra{\vb*{n}}\ket*{\Psi}=\bra{\vb*{n}}B(g)\ket*{\Psi}=B(D(g))_{\vb*{n},\vb*{n}}\bra{\sigma_g(\vb*{n})}\ket*{\Psi},
\end{equation} 
where Lemma~\ref{monomial} is applied.
Thus, $\abs{\bra{\vb*{n'}}\ket*{\Psi}}=\text{const.}$ for any $\ket*{\vb*{n'}}\in \mathcal{O}_{\vb*{n}}$.
\end{proof}
\begin{proof}[Proof of Corollary~\ref{interference}]
Because $\ket*{\Psi_\text{out}} \in V^{G'}_{\lambda'}$ from Thm.~\ref{main_thm}(ii), Thm.~\ref{main_thm}(iii) yields $\bra{\vb*{n}}\ket*{\Psi_\text{out}}=0$ when $\rho_\lambda^G\ket*{\vb*{n}}=0$.
This condition is equal to Eq.~\eqref{condition} from Thm.~\ref{main_thm}(i), when $\ket*{\Psi}=\ket*{\vb*{n}}$ and $G$ consists of diagonal matrices.
\end{proof}
\begin{proof}[Proof of Corollary~\ref{stab_meas_formula}]
Note that any Fock state is a joint eigenstate of $B(G')$ when $G'$ consists of diagonal matrices, as seen from Eq.~\eqref{permutation_formula}.
Then, the projector can be expressed as $\rho^{G'}_{\lambda'} = \sum_{\ket*{\vb*{n}}\in V^{G'}_{\lambda'}} \dyad{\vb*{n}}$.
From the definition, the probability of obtaining the measurement results $g=\lambda(g)$ for each $g\in G$ is 
\begin{equation}
    \sum_{\ket*{\vb*{n}}\in V_{\lambda'}^{G'}} \abs{\bra{\vb*{n}} B(U)\ket*{\Psi}}^2,
\end{equation}
which is equal to 
\begin{equation}
    \bra{\Psi} B(U^\dag) \rho^{G'}_{\lambda'} B(U)\ket*{\Psi}=\norm{\rho^{G}_{\lambda} \ket*{\Psi}}^2.
\end{equation}
\end{proof}

\subsection{B. Proof of Theorem~\ref{bell_thm}}
\begin{definition}\label{Dicke}
    The Dicke states are
    \begin{equation}
        \ket*{D^m_{k}} = \binom{m}{k}^{-\frac{1}{2}} \sum_{\substack{x\in \{0,1\}^{m}\\ wt(x)=k}} \ket*{x},
    \end{equation}
    where $wt(x)$ is a Hamming weight of $x$.
\end{definition}
\begin{lemma}\label{probability}
    Assume an Abelian transitive subgroup of symmetry group $\Sigma \subset S_m$, where there exists unique $\sigma \in \Sigma$ such that $\sigma(0)=i$ for each $i\in \{0,\cdots, m-1\}$.
    Letting ${P}_j(\theta)$ be the operator acting on the $j$th modes as ${P}_j(\theta)\ket*{{k}}_j=e^{ik\theta}\ket*{k}_j$, 
    \begin{equation}\label{prob_Dicke_formula}
        \norm*{\rho^{\Sigma}_1 {P}_0(\theta) \ket*{D^m_{k}}}^2 = \abs{\frac{(m-k)+e^{i\theta}k}{m}}^2,
    \end{equation}
    where $\rho^{\Sigma}_1$ is the projector onto the joint eigenstate of $\{B(P_\sigma)|\sigma \in \Sigma\}$ with eigenvalues of $1$.
\end{lemma}
\begin{proof}
From the definition, the Dicke states are invariant under any permutation of the modes; that is, 
\begin{equation}
    B(P_\sigma)\ket*{D^m_{k}}=\ket*{D^m_{k}}
\end{equation}
for any $\sigma \in S_m$.
Further, because ${P}_i(\theta)$ acts on the $i$th modes, 
\begin{equation}
    B(P_\sigma){P}_i(\theta)B(P_\sigma)^\dag={P}_{\sigma(i)}(\theta),
\end{equation}
from Eq.~\eqref{permutation_formula}.
From these equations and Lemma~\ref{projector}, 
\begin{equation}\label{cal1}
    \rho^{\Sigma}_1 {P}_0(\theta) \ket*{D^m_{k}} = \frac{1}{\abs{\Sigma}}\sum_{\sigma\in \Sigma} {P}_{\sigma(0)}(\theta) \ket*{D^m_{k}}.
\end{equation}
The group $\Sigma$ being an Abelian transitive group implies that there is only one element $\sigma \in \Sigma$ such that $\sigma(0)=i$ for each $i\in \{0,\cdots, m-1\}$, and that $\abs{\Sigma}=m$.
Therefore, Eq.~\eqref{cal1} is equivalent to
\begin{equation}
    \frac{1}{m}\sum_{i=0}^m {P}_i(\theta) \ket*{D^m_{k}} = \frac{(m-k)+e^{i\theta}k}{m}\ket*{D^m_{k}}.
\end{equation}
Therefore, Eq.~\eqref{prob_Dicke_formula} holds.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{bell_thm}]
From the main text, the remainder of the proof of Eq.~\eqref{measurment_basis} involves showing that the LOC consisting of $I_2 \otimes F_m$, $2m$ detectors, and ancilla photons $\ket*{\beta^-}^{\otimes m-1}$ corresponds to the measurement on $\ket*{\beta^\pm}$ expressed as
\begin{equation}\label{measurment_basis2}
    \{\sqrt{1-m^{-1}}\bra{\beta^+}, K'_0, \cdots, K'_m \},
\end{equation}
where 
\begin{equation}\label{Kraus'}
    K'_k = \sqrt{\binom{m}{k}}\frac{1}{2^{m/2}}\qty[\bra{\beta^-} + \frac{(m-2k)}{m}\bra{\beta^+}].
\end{equation}
Here, the LOC $I_2 \otimes F_m = F_m \oplus F_m$ preserves the number of photons incident on each block of the LOC. 
Thus, we can distinguish the input state according to these numbers.
For simplicity, we introduce the notation $\ket*{\overline{0}}=\ket*{20}$ and $\ket*{\overline{1}}=-\ket*{02}$.
Let $\overline{P}_i(\theta)$ be the operator acting as $\overline{P}_i(\theta)\ket*{x}_i=e^{ix\theta} \ket*{x}_i$ for $x=\{\overline{0},\overline{1}\}$. 
Then, we can expand the state as
\begin{equation}
    \ket*{\beta^+}\ket*{\beta^-}^{\otimes m-1} = \frac{1}{2^{m/2}}\overline{P}_0(\pi)(\ket*{\overline{0}}+\ket*{\overline{1}})^{\otimes m} 
    =\frac{1}{2^{m/2}} \sum_{k=0}^{m}\binom{m}{k}^{\frac{1}{2}} \overline{P}_0(\pi) \ket*{\overline{D}_k^m},
    \label{expansion}
\end{equation}
where $\ket*{\overline{D}_k^m}$ is the Dicke state in Def.~\ref{Dicke} defined for $\{\ket*{\overline{0}},\ket*{\overline{1}}\}^{\otimes m}$.
The number $k$ in Eq.~\eqref{expansion} is obtained from the measurement result.
The LOC $I \otimes X_m$ corresponds to the permutation of $(0\cdots m-1)$ of the modes of $\{\ket*{\overline{0}},\ket*{\overline{1}}\}^{\otimes m}$.
Thus, Lemma~\ref{probability} yields 
\begin{equation}\label{probability_cal}
    \norm*{\rho^{G}_1 {\overline{P}}_0(\pi) \ket*{\overline{D}^m_{k}}}^2 = \norm*{\rho^{C_n}_1 {P}_0(\pi) \ket*{D^m_{k}}}^2 = \qty\Big(\frac{m-2k}{m})^2,
\end{equation}
where $G=\langle I \otimes X_n\rangle$ and $C_m = \langle (0 \cdots m-1)\rangle$.
Because $\ket*{\beta^-}^{\otimes m} \in V_1^G$, we can distinguish $\ket*{\beta^+}$ from $\ket*{\beta^-}$ when we measure $I \otimes X_n \neq 1$, the total probability of which is
\begin{equation}
    \frac{1}{2^m}\sum_{k=0}^{m}\binom{m}{k}\qty\Big(1-\qty\Big(\frac{m-2k}{m})^2)=1-\frac{1}{m}.
\end{equation}
In contrast, when we obtain $I \otimes X_n = 1$, we can expect the corresponding Kraus operators to be Eq.~\eqref{Kraus'}, from Eq.~\eqref{probability_cal}.
To completely identify the measurement basis, we need to perform a kind of measurement tomography, in which we consider multiple bases as the input states and reconstruct the measurement corresponding to this setup.
For input states
\begin{equation}
    \frac{1}{\sqrt{2}}(\ket*{\beta^-}_0\pm i \ket*{\beta^+}_0) = \frac{e^{\pm i \pi/4}}{\sqrt{2}} P_0(\mp \frac{\pi}{2})_0 (\ket*{\overline{0}}+\ket*{\overline{1}}),
\end{equation}
We can calculate the measurement probabilities from Lemma~\ref{probability} in the same manner.
Therefore, we can ensure that the considered setup corresponds to the measurement represented by Eq.~\eqref{measurment_basis2}.
Thus, the entire setup corresponds to the quantum measurement on two qubits represented by Eq.~\eqref{measurment_basis}.
\end{proof}
Following Ref.~\cite{Kim2022-oj}, we introduce the relative entropy of entanglement of quantum measurements as an entanglement measure. 
Computing this measure is generally difficult; however, we can explicitly calculate it in the special case as follows.
\begin{definition}\label{measure_def}
    For the $d$-dimensional systems $A$ and $B$, consider a quantum measurement on the composite system $AB$, which is described by positive operator-valued measure (POVM)  elements as $M_{AB}=\{M_x\}$.
    The relative entropy of entanglement of a quantum measurement $M_{AB}$ is
    \begin{equation}
        E(M_{AB})=\frac{1}{d^2} \min_{F_{AB} \in \text{sepM(A:B)}}\sum_x S(M_x||F_x),
    \end{equation}
    where $S( \cdot || \cdot )$ is the quantum relative entropy and $\text{sepM(A:B)}$ is the set of separable measurements on $A$ and $B$~\cite{Watrous2018-la}.
\end{definition}
\begin{lemma}\label{RE_lemma}
    Assume that the rank of all POVM elements is $1$ in Def.~\ref{measure_def}. Then, 
    \begin{equation}\label{RE_formula}
        E(M_{AB}) = \frac{1}{d^2} \sum tr(M_x) E_s \qty\Big(\frac{M_x}{\tr(M_x)}),
    \end{equation}
    where $E_s(\cdot)$ is the entanglement entropy of quantum states.
\end{lemma}
\begin{proof}
    Let $M_x =p_x \rho_x$ with $p_x=\tr(M_x)$, and $F_x=q_x \sigma_x$ with $q_x=\tr(F_x)$.
    Then, 
    \begin{align}\label{RE_calculation}
        E(M_{AB}) 
        &= \frac{1}{d^2} \min_{F_{AB} \in \text{sepM(A:B)}}\sum_x \qty\Big{p_x S(\rho_x||\sigma_x) + p_x \log{\frac{p_x}{q_x}}} \notag \\
        &=\frac{1}{d^2} \min_{F_{AB} \in \text{sepM(A:B)}} \qty\Big{\sum_x p_x S(\rho_x||\sigma_x) + S(p||q)},
    \end{align}
    where $S(p||q)$ is the classical relative entropy of $p=\{p_x\}$ with respect to $q=\{q_x\}$.
    $S(p||q)$ takes the minimum value of $0$ when $q=p$. 
Here, $\rho_x$ is a pure state, and $S(\rho_x||\sigma_x)$ takes the minimum value of $E_s(\rho_x)$~\cite{Vedral1998-fg}.
    Letting $\mathcal{I}$ and $\Delta$ be the identity and completely dephasing channel for any fixed basis, respectively, the state at the minimum is expressed as $\sigma_x=(\mathcal{I} \otimes \Delta)(\rho_x)$.
    Letting $F_{AB}=\{F_x\}=\{p_i (\mathcal{I} \otimes \Delta)(\rho_x)\}_x$, $F_{AB}$ satisfies the conditions for being a quantum measurement, especially as
    \begin{equation}
        \sum_x F_x = \sum_x p_x (\mathcal{I} \otimes \Delta)(\rho_x) = (\mathcal{I} \otimes \Delta)(I) = I,
    \end{equation}
    from $\sum_x p_x \rho_x = \sum_x M_x = I$.
    Thus, the quantum measurement $F_{AB}$ simultaneously minimizes each term in Eq.~\eqref{RE_calculation} and, hence, Eq.~\eqref{RE_formula} holds.
\end{proof}

We use the following Lemma~\ref{known_thm} without proof.
\begin{lemma}\label{known_thm}
    For the binary entropy function $H_b$, 
    \begin{equation}\label{inquality1}
        4p(1-p) \leq H_b(p) \leq (4p(1-p))^{1/ln4}
    \end{equation}
    for $0 \leq p \leq 1$~\cite{Topsoe2001-wd}, and 
    \begin{equation}\label{inquality2}
        \sqrt{\frac{n}{8 k (n-k)}}  < \binom{n}{k}2^{-nH_b(k/n)} \leq  \sqrt{\frac{n}{\pi k (n-k)}} 
    \end{equation}
    for integers $n$ and $1 \leq  k \leq n-1$~\cite{Cover2006-eh}.
\end{lemma}
\begin{lemma}\label{F_function_lemma}
    Let 
    \begin{equation}
        F(x)=-x^2\log_2{\frac{x^2}{x^2+(1-x)^2}}-(1-x)^2\log_2{\frac{(1-x)^2}{x^2+(1-x)^2}}
    \end{equation}
    and 
    \begin{equation}\label{f_function_def}
        f(x)=\frac{4x^2(1-x)^2}{x^2+(1-x)^2}.
    \end{equation}
    Then,  
    \begin{equation}\label{inequality2}
        f(x) \leq F(x) \leq \frac{1}{2}
    \end{equation}
    for $0\leq x \leq 1$ and $f'(x) \geq 0$ for $0\leq x \leq \frac{1}{2}$~\footnote{We could not find the proof of $F'(x) \geq 0$. Therefore, we introduced the function $f$ and proved $f'(x) \geq 0$ instead.}.
\end{lemma}
\begin{proof}
With the binary entropy function $H_b(x)$, $F$ can be expressed as
\begin{equation}
    F(x)=(x^2+(1-x)^2)H_b\qty\Big(\frac{x^2}{x^2+(1-x)^2}).
\end{equation}
For $0\leq x \leq 1$, it holds that  $(4x(1-x))^{1/ln4}\leq 2\sqrt{x(1-x)}$.
Thus, Eq.~\eqref{inquality1} in Lemma~\ref{known_thm} yields the following inequality:
\begin{equation}\label{F_ub_formula}
    f(x) \leq F(x) \leq 2x(1-x) \leq \frac{1}{2}.
\end{equation}
The first derivative of $f$ is
\begin{equation}
    f'(x)=\frac{8x(1-x)((1-x)-x)(1-x(1-x))}{(x^2+(1-x)^2)^2} \geq 0
\end{equation}
for $0\leq x \leq \frac{1}{2}$.
\end{proof}
\begin{proof}[Proof of Corollary~\ref{bell_values}]
The average success probability $P_m$  of Bell-state discrimination is calculated as the probability that the maximally mixed input state is projected to one of the Bell states.
For even $m$, the case of $k=\frac{m}{2}$ is special because the corresponding Kraus operator becomes the projection onto a Bell state $\ket*{\phi^-}$, with probability $\binom{m}{m/2}2^{-m}$.
Therefore, $P_m$ is calculated as in Eq.~\eqref{success_probability} in total.
The relative entropy of entanglement $E_m$ can be calculated as in Eq.~\eqref{entanglement_measurement} using Lemma~\ref{RE_lemma}.

From Eq.~\eqref{inquality1} in Lemma~\ref{known_thm}, $P_m$ are upper bounded as
\begin{equation}
    P_m \leq P_u(m) = \frac{1}{4}\qty(3-\frac{1}{m}+\sqrt{\frac{4}{\pi m}}).
\end{equation}
From the calculation, $P_u'(x)<0$ for $x>\pi$. 
Numerically, we confirmed that $P_u(44) \simeq 0.7868$ and that the maximum value of $P_m$ for $m<44$ is 
\begin{equation}
    P_8 = \frac{403}{512} \simeq 0.7871.
\end{equation}
Thus, $P_8$ is the maximum value of $P_m$, from $P_8>P_u(44)$.
Moreover, $P_m \rightarrow \frac{3}{4}$ for $m \rightarrow \infty$ are proved by lower-bounding $P_m$ in the same manner.

Letting 
\begin{equation}
    S(m) = \frac{1}{2^{m-1}}\sum_{k=0}^m \binom{m}{k}F\qty\Big(\frac{k}{m}),
\end{equation}
$S(m)\rightarrow 1$ yields $E_m \rightarrow 1$ for $m\rightarrow \infty$. 
Because $F(x)\leq \frac{1}{2}$ for $0\leq x \leq 1$, from Lemma~\ref{F_function_lemma}, it holds that
\begin{equation}
    S(m) \leq \frac{1}{2^m}\sum_{k=0}^m \binom{m}{k} = 1.
\end{equation}
Introducing a real number $0< c < \frac{1}{2}$, we obtain the lower bound of $S_m$ as
\begin{equation}
    S _m > \frac{1}{2^{m-1}} \sum_{k=\lceil cm \rceil}^{m-\lceil cm \rceil} \binom{m}{k}F\qty\Big(\frac{k}{m}) \\
    > f(c) \frac{1}{2^{m-1}} \sum_{k=\lceil cm \rceil}^{m-\lceil cm \rceil} \binom{m}{k},
\end{equation}
where $f$ is defined in Eq.~\eqref{f_function_def}, and the second inequality holds from $F(x) \geq f(x)$ and $f'(x) \geq 0$ for $0\leq x< \frac{1}{2}$ in Lemma~\ref{F_function_lemma}.
Further, Eq.~\eqref{inquality1} in Lemma~\ref{known_thm} implies the inequality 
\begin{equation}
    \binom{m}{k} \leq 2^{m H_b(\frac{k}{m})}
\end{equation}
for $0\leq k \leq n$. 
Thus, 
\begin{equation}
    \sum_{k=0}^{\lceil cm \rceil-1} \binom{m}{k}  \leq c m 2^{m H_b(c)}.
\end{equation}
Finally, we have the following lower bound: 
\begin{equation}\label{lower_bound}
    S(m) > 2f(c)(1-2cm 2^{-m(1-H_b(c))}).
\end{equation}
It holds that  $H_b(c)<1$ for $c\neq \frac{1}{2}$ and $f(\frac{1}{2})=\frac{1}{2}$.  
Thus, Eq.~\eqref{lower_bound} implies that $S(m)\rightarrow 1$ for $m\rightarrow \infty$ and $c\rightarrow \frac{1}{2}$.
\end{proof}

