Cardinality of $\text{Aut}(G\times G) $
Solution 1:
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.