I am new to Algebraic Graph Theory and currently reading Algebraic Graph Theory by Godsil and Royle. In page $5,$ the author provided the following lemma and its proof:
Lemma 1.3.1 If $x$ is a vertex of the graph $X$ and $g$ is an automorphism of $X,$ then the vertex $y = x^g$ has the same valency as $x.$
Proof: Let $N(x)$ denote the subgraph of $X$ induced by the neighbors of $x$ in $X.$ Then $$N(x)^g = N(x^g) = N(y),$$ and therefore $N(x)$ and $N(y)$ are isomorphic subgraphs of $X.$ Consequently, they have the same number of vertices, and so $x$ and $y$ have the same valency.
Question: Intuitively, I understand that $N(x)^g = N(x^g).$ However, I am unable to prove it rigorously. Any hint would be appreciated.
Notation: The valency of a vertex $x$ in a graph $X$ is the number of neighbors of $x.$
Fix two vertices $x$ and $y$ in a graph $X.$ We denote $x\sim y$ if there is an edge between $x$ and $y.$ If $x\sim y,$ then we say that $y$ is a neighbors of $x.$
Two graphs $X$ and $Y$ are isomorphic if there exists a bijection $\phi$ which maps set of vertices of $X$ onto set of vertices of $Y$ such that $x\sim y$ if and only if $\phi(x)\sim\phi(y).$
We say that $g$ is an automorphism of $X$ if $g$ maps set of vertices of $X$ onto itself.
If $x$ is an vertex of $X,$ then we denote image of $x$ under $g$ as $x^g.$
Say that $N(x) = \{x_1, x_2, \dots, x_k\}$. So $x \sim x_i$ for $i=1, \dots, k$. Since $g$ is an isomorphism, we have that $g(x) \sim g(x_i)$ or, equivalently, $x^g \sim x_i^g$, for $i=1, \dots, k$. Therefore $N(x^g)=N(x)^g$.