Elements of $\operatorname{Aut}(\operatorname{Aut}(G))$ acting as an identity on $\operatorname{Inn}(G)$, for $G$ a nonabelian, simple group

Question

Let $f$ be an element of $\operatorname{Aut}(\operatorname{Aut} G)$ s.t $f(r)$ is equal to $r$ for all $r \in \operatorname{Inn}(G)$. Prove that $f(j)$ is equal to $j$, $\forall f\in\operatorname{Aut}(G)$ where G is non-abelian and simple.

I have taken an element of $\operatorname{Aut}(G)$, say $h$ then $h$ is equal to $g\cdot k$ where $g\in \operatorname{Aut}(G)$ and $k\in \operatorname{Inn}(G)$. Next what?

Answer 1

Let $I ={\rm Inn}(G) \cong G$, $A = {\rm Aut}(G)$ and $B = {\rm Aut}(A)$. Since $Z(A)=1$, we have $A \cong {\rm Inn}(A)$, so we can identify $A$ with the normal subgroup ${\rm Inn}(A)$ of $B$.

Now $C:=C_B(I)$ is also a normal subgroup of $B$ and so $[C,A] \le C \cap A = 1$, because no nontrivial element of $A$ centralizes $I \cong G$. So $C$ centralizes $A$ i.e. $C$ acts as the identity on ${\rm Aut}(G)$.