Answers from the comments (hence CW):
> This is only true if $\textrm{Aut}(G)$ is a complete group; that is, $\textrm{Aut}(G) \cong \textrm{Aut}(\textrm{Aut}(G))$ via the canonical homorphism $Aut(G)\to Aut(Aut(G))$. In particular, $G=S_{3} \times S_{3}$ is a counterexample. – Steve D
> @SteveD In fact it is true if and only if $\textrm{Aut}(G)$ is complete. – Derek Holt