TITLE: Does every automorphism of G come from an inner automorphism of S_G?
QUESTION [12 upvotes]: I feel sort of silly asking this question. Unless I'm very much mistaken the paper I'm reading assumes the following statement:
Let $G$ be a finite group. We may embed it via the Cayley embedding into an ambient permutation group $G \leq S_{|G|}$. Then any automorphism of $G$ comes from conjugation by an element in $N_{S_{|G|}}(G)$.
Is this statement true?

REPLY [16 votes]: The statement is true. Let $g \in G$ and $\pi \in Aut(G)$. Let $\lambda_{g}$ be the corresponding left translation by $g$. Regard $\pi$ and $\lambda_{g}$ as elements of $Sym(G)$. Then for all $x \in G$,
$(\pi \lambda_{g} \pi^{-1})(x) = (\pi \lambda_{g}) ( \pi^{-1}(x)) = 
\pi( g \pi^{-1}(x)) = \pi(g) x = \lambda_{\pi(g)}(x)$.
Thus $\pi \lambda_{g} \pi^{-1} = \lambda_{\pi(g)}$.