I have a question about group $G$ which satisfies Inn$(G) $ char Aut$(G)$ and $Z(G)$=$\{1\}$. Let $G$ be a group which satisfies $Z(G)=\{1\}$ and ${\rm Inn(G)} \space \mathbb{char} \space {\rm Aut(G)}$; then every automorphism of $A={\rm Aut(G)}$ is an inner automorphism. ($H \space \mathbb{char} \space G$ means that $H$ is a characteristic subgroup of $G$. Note that we can assume $G \subseteq A$, since $Z(G)=\{1\}$ so $G \cong{\rm  Inn}(G)$.)

I am given a hint, and it says that $C({\rm Inn}(G)) \unlhd {\rm Aut}(A)$, so one derives $C({\rm Inn}(G)) \subseteq C(A)$. I'm stuck only on this point. Why can we say that? ($C(H)$ means the centralizer of $H$ in ${\rm Aut}(A)$.)
Let $I = {\rm Inn}\, G$ and $C = C_{{\rm Aut}(A)}(I)$. Then $C \unlhd {\rm Aut}(A)$ and $A \unlhd {\rm Aut}(A)$. Since $A$ is by definition the group of automorphisms of $G \cong I$, no nontrivial element of $A$ can centralize $I$; i.e. $C \cap A = 1$. Hence $[C,A] \le C \cap A = 1$; i.e. $C \le C(A)$.
I post my solution here:(to prove $C_{AutA}(I)=1$)
http://www.artofproblemsolving.com/community/c7t435f7h622064_automorphism_group_of_automorphism_group_of_simple_group_g