Let $G$ a non-abelian simple group and let $A=\mathrm{Inn}(G)$ and $B=\mathrm{Aut}(G)$. I would like to know the solution to $A\,\operatorname{char}\, B$. However, I know the following. Let $\phi \in \mathrm{Aut}(B)$.
a) $\phi(A) \cong A \cong G/Z(G) \cong G$ is simple.
b) $A \cap \phi(A)$ is normal in $\phi(A)$.
c) Since a) and b), $A \cap \phi(A) = \{ \rm{id}_G \}$ or $\phi(A)$.
Therefore, I only want to know $A \cap \phi(A) \neq \{ \rm{id}_G \}$.
$\def\Inn{\operatorname{Inn}}$$\def\Aut{\operatorname{Aut}}$Let $G$ be a group with $Z(G) = 1$. Write $i_g$ for the inner automorphism $i_g(x) = gxg^{-1}$, and observe that $\phi i_g \phi^{-1} = i_{\phi(g)}$ for any $g \in G$ and $\phi \in\Aut(G)$. Therefore if $\phi$ commutes with each $i_g \in \Inn(G)$, i.e., $\phi \in C_{\Aut(G)}(\Inn(G))$, it follows that $i_g = i_{\phi(g)}$ for all $g \in G$, whence $\phi(g) = g$ since $Z(G)$ is trivial. Therefore the centralizer $C_{\Aut(G)}(\Inn(G))$ is trivial.
Now assume $G$ is simple. We claim that $\Inn(G) \cong G$ is characteristic in $\Aut(G)$. Let $\psi \in \Aut(\Aut(G))$, let $I = \Inn(G)$, and let $J = \psi(\Inn(G))$. Then $I \cong J \cong G$ and $I, J \trianglelefteq \Aut(G)$, so $I \cap J$ is either $I$ or trivial (as explained by OP). In the latter case we have $[I,J] \le I \cap J = 1$, so $J \le C_{\Aut(G)}(I) = 1$, a contradiction.
Bonus: It follows that we have a natural map $\Aut(\Aut(G)) \to \Aut(\Inn(G)) \cong \Aut(G)$ defined by restriction. The first paragraph above shows that (a) this map is injective, since its kernel is $C_{\Inn(G)}(\Aut(G)) = 1$, and (b) this map is surjective, since the inner automorphism $i_\phi \in \Inn(\Aut(G))$ maps to $\phi \in \Aut(G)$. Therefore $\Aut(\Aut(G)) = \Inn(\Aut(G)) \cong \Aut(G)$.
This is all essentially abstract nonsense and it is not necessary to use any deep facts about finite simple groups such as Schreier's hypothesis. In fact we do not need to assume $G$ is finite.
A famous conjecture of Otto Schreier states that for every finite simple group, Aut$(G)$/Inn$(G)$ (a.k.a. Out$(G)$) is always solvable. The conjecture was proved to be true as a consequence of the classification of finite simple groups. Therefore Inn$(G)$ is in fact the unique subgroup isomorphic to $G$ inside Aut$(G)$, whence characteristic: if $H$ would be another subgroup of Aut$(G)$ isomorphic to, but different from Inn$(G)$, then, since $H$ is non-abelian simple, and Inn$(G) \unlhd $ Aut$(G)$, $H \cap $Inn$(G)=1$. But then $H \cong H/H \cap $Inn$(G) \cong H \cdot $Inn$(G)/$Inn$(G)$ is solvable, a contradiction.
