Automorphism group of a Cayley graph

Question

Let $G$ be a group. Let $\Gamma = \Gamma(G,X)$ be the Cayley graph of $G$ defined with respect to a generating set $X$. I want to show that $G\cong \text{Aut}(\Gamma)$. Note that by $\text{Aut}(\Gamma)$ I am not referring to the automorphism group of the underlying undirected graph, but rather the detailed graph where each edge is directed and labelled with the appropriate generator.

For example, in the following directed and labelled graph, there is only one non-trivial automorphism: the one where I send $1$ to $4$. Indeed, the rest of the automorphism is uniquely determined by describing the image of a single vertex under the automorphism.

I tried to follow this post, but I was a little confused. My questions are the following:

1. How are the elements of $\text{Aut}(\Gamma)$ defined? Since it is different from the usual definition of a graph isomorphism, I wasn't sure how to go about making this definition.
2. Why is it easy to see that $T_h\in\text{Aut}(\Gamma)$? (I suppose the answer to this question depends on how $\text{Aut}(\Gamma)$ is defined.)

Answer 1

The vertices of the Cayley graph $\Gamma$ are the elements of $G$, there is an edge $(g,gs)$ for every $s\in X$ (where $X$ is the generating set) and $g\in G$. The edge $(g,gs)$ is labelled by the generator $s$. An automorphism of $\Gamma$ is a permutation of the vertex set which induces a permutation of the edge set that also preserves the labels of the edges. (It sends any edge to an edge with the same label.)

Call such a function $\phi:G\to G$. The fact it preserves edges means $(\phi(g),\phi(gs))$ must be an edge for all $g\in G,s\in S$, and it must have the same label $s$, which means the second coordinate $\phi(gs)$ must be the first one, $\phi(g)$, times $s$. That is, $\phi(gs)=\phi(g)s$ for all elements $g\in G,s\in S$. You can do the same idea for the edges $(gs^{-1},g)$ labelled by $s$ in order to show $\phi(gs^{-1})=\phi(g)s^{-1}$ too.

Since all $g\in G$ are products of elements of $S$ and their inverses, by induction $\phi(g)=\phi(e)g$. That is, any labelled automorphism $\phi$ of $\Gamma$ is simply left multiplication $T_h(g):=hg$ by a group element $h=\phi(e)$, and conversely (which follows simply from the associative property).