Automorphism group of labelled Cayley graph

Question

In the proof of the following statement:

The automorphism group of labelled Cayley graph of $G$ is $G$ itself

given here: Prove that the group of automorphisms of a labelled Cayley graph of a group G is the group G itself (Just stumped on one direction)

Part of the proof says: " This subgroup acts regularly (i.e. transitively and with trivial stabilizers) on $\Gamma$, so to prove that it is equal to $Aut(G)$, it is sufficient to prove that the stabilizer in $Aut(G)$ of a vertex in $\Gamma$ is trivial. "

My question is why is this condition sufficient? That is why if the group acts regularly on $\Gamma$ and it is isomorphic to a subgroup of $Aut(\Gamma)$, then it is isomorphic to $Aut(\Gamma)$?

(And also I assume $Aut(G)$ in the proof should change to $Aut(\Gamma)$?)