What is ${\rm Aut}(G^n)$? Suppose $G$ a finite indecomposable group such that $Z(G)=\{e\}$.
Now consider $G^n$. What is ${\rm Aut}(G^n)$?
My claim is ${\rm Aut}(G^n)={\rm Aut}(G)^n \rtimes S_n$. 
 A: The Krull-Schmidt Theorem (see Wikipedia article - it was apparently first proved by Wedderburn, and also by Remak) says that for a group $G$ satisfying ACC and DCC on normal subgroups (which holds in particular for finite groups), if $G \cong G_1 \times \cdots \times G_k \cong H_1 \times \cdots \times H_l$, with each $G_i,H_i$ indecomposable, then $k=l$, the $H_i$ can be reordered such that $G_i \cong H_i$ for all $i$, and also $G \cong G_1 \times \cdots \times G_{k-1} \times H_k$. In other words, $G_k$ can be replaced by $H_k$ in the decomposition.
Now assume that $Z(G)=1$. Since $G_k$ can be replaced by $H_k$, $H_k$ must project onto the quotient group $G/(G_1 \times \cdots \times G_{k-1})$. Let $g =(g_1,\ldots,g_{k-1},g_k) \in H_k$ with $g_i \in G_i$. If some $1 \ne g_i$ with $i \le k-1$ then, since $g_i \not\in Z(G_i)$, taking the commutator of $g$ with an element of $G_i$ that does not centralize $g_i$ gives an element in $G_i \cap H_k$, contradiction. So $G_k=H_k$. Similarly we can show that $G_i=H_i$ for all $i$, so the direct decomposition of $G$ into indecomposable groups is unique, and not just unique up to isomorphism.
So, if $Z(G) = 1$ and $G \cong G_1 \times \cdots \times G_k$ with $G_i$ indecomposable, then ${\rm Aut(G)}$ is isomorphic to the semidirect product of ${\rm Aut}(G_1) \times \cdots \times {\rm Aut}(G_k)$ with the subgroup of $S_k$ that permutes the isomorphic $G_i$.
