Let $G$ be a group. Now consider $\operatorname{Aut} G$ then $\operatorname{Aut} (\operatorname{Aut} G) = \operatorname{Aut}^2(G)$, then $\operatorname{Aut}(\operatorname{Aut}(\operatorname{Aut} G))=\operatorname{Aut}^3(G)$ and so on.
Let $n$ be the least positive integer for which $\operatorname{Aut}^n (G)$ is isomorphic to $\operatorname{Aut}^{n+1}(G)$. Find the value of $n$ for the groups $\Bbb Z_4$, $\Bbb Z_3$, $\Bbb Z_2 \times \Bbb Z_2$ and explain why it is so. Just use Automorphism in Group theory.
For $\Bbb Z_4$ I can send the generator $1$ of $\Bbb Z_4$ to $1$ or $3$. So Aut($\Bbb Z_4$)=$\Bbb Z_2$ . Again $Aut(\Bbb Z_2)=\Bbb Z_2$. So we are done. Similarly for $\Bbb Z_3$. Now what about $\Bbb Z_2 \times \Bbb Z_2$?
I guess by $\mathbb{Z}_n$ ($n$-adic numbers) the author actually means $\mathbb{Z}/n\mathbb{Z}$ or short $\mathbb{Z}/n$ (the notation $\mathbb{Z}_n$ for this is common, but still wrong.)
One checks directly that every permutation of the non-zero elements of $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ is actually an automorphism. It follows $\mathrm{Aut}(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}) \cong S_3$. It is well-known that $\mathrm{Aut}(S_n) \cong S_n$ if $n \neq 6$.