in $S_n$ can act on a vector $\pmb{u}=(u_1,\ldots,u_n)^t$ of $\operatorname{\mathbb{R}}^n$ in a natural way: $$\label{Def-PermutationOperator} \sigma \hspace{0.5mm} \pmb{u}=(u_{\sigma^{-1}1},u_{\sigma^{-1}2},\ldots,u_{\sigma^{-1} n})^t,$$ where $\operatorname{\mathbb{R}}^n$ is the $n$-dimensional vector space over the real field $\operatorname{\mathbb{R}}$. Accordingly, any permutation $\sigma$ in $S_n$ can be regarded, through the action on vectors, as a linear operator on $\operatorname{\mathbb{R}}^n$. We call a (0,1)-square matrix a [*permutation matrix*]{} if in each row and column there is exactly one entry that is equal to 1. It is easy to check that the matrix $\mathbf{P}_{\sigma}$ of the operator $\sigma$ with respect to the standard basis $\pmb{e}_1,\ldots,\pmb{e}_n$ is a permutation matrix, where each $\pmb{e}_i$ $(i=1,\ldots,n)$ has exactly one non-trivial entry on $i$th coordinate that is equal to 1, and all other entries of $\pmb{e}_i$ are equal to 0.

Recall that the vertex set $V(G)$ is $[n]$, so a bijective map $\phi$ from $V(G)$ to itself is a permutation of $[n]$. It is easy to check that $$\label{Equation00} 
\phi\mbox{ is an automorphism of }G\mbox{ if and only if }\mathbf{P}_{\phi}^{-1}{\mathbf{A}( G )}\mathbf{P}_{\phi}={\mathbf{A}( G )},$$ which presents an algebraic way of characterizing automorphisms of $G$. There is in virtue of eigenspaces of ${\mathbf{A}( G )}$ another way to characterize automorphisms of $G$.

\[Lem-AutomorphismAndOperator\] Let $G$ 