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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 $Aut(\Gamma)$?)

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    Are you familiar with the orbit-stabilizer theorem? – 2017-02-01
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    Yes, how do you apply it here, assuming we have a single orbit? – 2017-02-01
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    If the stabilizer is trivial then the order of the group equals the size of the orbit. – 2017-02-01
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    Thanks, now it's clicking. So a different way of saying this is: If the group $G$ acts regularly on a set, then since the stablizer of "a" member of the set $\Gamma$ is trivial, the size of the group is equal to the size of the set, and if the group $G$ is isomorphic to a subgroup of $Aut(\Gamma)$ then it follows that $G \cong Aut(\Gamma)$. Is this entirely correct? – 2017-02-01
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    Seems ok, but I don't know the details about $\Gamma$. – 2017-02-01

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