TITLE: Cardinality of $\text{Aut}(G\times G) $
QUESTION [11 upvotes]: Let $G$ be a finite group. If $|\text{Aut}(G)|$ is known, what can we say about $|\text{Aut}(G\times G)|$ ?

REPLY [10 votes]: Picking up on Mike Miller's comment, we can do slightly better for a lower bound:
$$\left|\operatorname{Aut}(G\times G)\right|\geq 2\left|\operatorname{Aut}(G)\right|^2.$$
Besides the disjoint copies of $\operatorname{Aut}(G)$ in  $\operatorname{Aut}(G\times G)$, we also have the automorphism that swaps the two copies of $G$.  Furthermore, for $G\neq 1$, this is the best possible lower bound, as $G = S_{3}$ (symmetric group of degree $3$) shows.