Inn characteristic in Aut If $G$ is a centerless group then is $\mathrm{Inn}(G)$ necessarily characteristic in $\mathrm{Aut}(G)$? 
The condition of being centerless is necessary as $D_8$ provides a counterexample otherwise.
 A: 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

A: Some more on the question.
If $Aut(G)$ is complete, then $Aut(Aut(G))=Aut(G)$, and then the only automorphisms $Inn(G)$ has to worry about are conjugations; it is thus characteristic.
If $Inn(G)$ is characteristic, then it is normal in $Aut(Aut(G))$.  If there exists $\beta\in Aut(Aut(G))\setminus Aut(G)$, then $\beta$ acts on $Inn(G)$ by conjugation, inducing an automorphism.  Thus there is an $\alpha\in Aut(G)$ such that $\beta^{-1}\alpha$ acts as the identity on $Inn(G)=G$. But then $\beta^{-1}\alpha\beta=\beta$ is in $Aut(G)$, a contradiction.  Thus $Aut(Aut(G))=Aut(G)$.
Finally, to see $G=S_3\times S_3$ is a counterexample, simply note that $Aut(G)=(S_3\times S_3)\rtimes C_2$, where the outer automorphism swaps the two factors. Now there are two copies of $S_3\times S_3$ in $Aut(G)$: the obvious one (the inner automorphisms), and the not-so-obvious one generated by the usual $3$-cycles, and the "transpositions" given by the outer automorphism, and a product of involutions in each $S_3$ factor that commutes with this outer automorphism.  These two copies of $S_3\times S_3$ are permuted by $Aut(Aut(G))$ (this permutation is the outer automorphism of $Aut(G)$).
