Suppose that $S$ is a non-abelian finite simple group with $\operatorname{Out}(S)>1$. The question is this: are there any non-abelian finite groups $G$ such that $\operatorname{Aut}(G)=S$?
I am imposing the condition that $G$ be non-abelian so as to exclude having the obvious $\operatorname{Aut}(C_2^n) \cong \operatorname{GL}_n(2)$, which is simple for $n>2$. The condition $\operatorname{Out}(S)>1$ implies that $G$ cannot be $S$ itself.
Clearly, $G/Z(G) \cong \operatorname{Inn}(G)$ is a normal subgroup of $S$, so either it is trivial (which is not allowed since $G$ is non-abelian) or $S$ itself. So a candidate $G$ must have the form $G=Z(G).S$ with $Z(G)>1$ (this notation expresses the fact that $G$ must be a non-split extension of $Z(G)$ by $S$). Thus the Schur multiplier of $S$ must be non-trivial.