Let $S_k \wr S_n$ be the wreath product of two symmetric groups (so $S_n$ acts on $(S_k)^n = S_k \times ... \times S_k$ by permuting the factors; we then take the semidirect product).
What is $Aut(S_k \wr S_n)?$
Let $S_k \wr S_n$ be the wreath product of two symmetric groups (so $S_n$ acts on $(S_k)^n = S_k \times ... \times S_k$ by permuting the factors; we then take the semidirect product). What is $Aut(S_k \wr S_n)?$ 


Let $G = S_k \wr S_n$. For $k \ge 5$, $G$ has the unique minimal normal subgroup $A_k^n$, which is therefore fixed by every automorphism $\phi$ of $G$. If $\phi$ centralizes $A_k^n$ then, for any $g \in G$, $h \in A_k^n$, we have $g^{1}hg = \phi(g^{1}hg) = \phi(g)^{1}h\phi(g)$. So $\phi(g)g^{1} \in G$ centralizes $A_5^n$, and hence $\phi(g)=g$, so $\phi=1$. Hence ${\rm Aut}(G) \le {\rm Aut}(A_k^n)$. For $k \ne 6$, we have ${\rm Aut}(A_k^n) \cong G$, so $G$ is complete. If $k=6$, then it is not hard to see that ${\rm Out}(G) = 2$, where the outer automorphism of order 2 acts as the same outer automorphism of $S_6$ on each of $n$ factors of $S_6^n$. I believe that $G$ is also complete when $k=4$ and when $k=3$ and $n>2$, although I have not tried to write down a proof. For $k=3$ and $n=2$ there is an outer automorphism of order 2. I have not thought at all about $k=2$, but note that $G$ has centre of order 2 in that case. 

