Is $\operatorname{Aut}(\mathbb{Z}_3\times \mathbb{Z}_3)$ isomorphic with $\operatorname{Aut}(\mathbb{Z}_3)\times \operatorname{Aut}(\mathbb{Z}_3)$? Is  $\operatorname{Aut}(\mathbb{Z}_3\times \mathbb{Z}_3)$ isomorphic with $\operatorname{Aut}(\mathbb{Z}_3)\times \operatorname{Aut}\left(\mathbb{Z}_3\right)$?
From the lecture I know that $\operatorname{Aut}\left(G\times H\right)\simeq \operatorname{Aut}\left(G\right)\times \operatorname{Aut}\left(H\right)$ if gcd$(G,H)=1$ but I don't know what I can use when $G=H$ so $\gcd(G,H)\neq1$.
Can I have any tips?
 A: For every integer $n \geq 2$, the homomorphisms of $\mathbf Z/(n) \times \mathbf Z/(n)$ to itself are the linear maps given by $2 \times 2$ mod $n$ matrices $A = (\begin{smallmatrix}a&b\\c&d\end{smallmatrix})$, which act on $\mathbf Z/(n) \times \mathbf Z/(n)$ by $A\binom{x}{y} = \binom{ax+cy}{bx+dy}$.  This homomorphism is invertible if and only if the matrix is invertible, which means $\det A$ is invertible mod $n$. Such $A$ form the group ${\rm GL}_2(\mathbf Z/(n))$ of invertible $2 \times 2$ mod $n$ matrices. So ${\rm Aut}(\mathbf Z/(n) \times \mathbf Z/(n)) = {\rm GL}_2(\mathbf Z/(n))$.
By similar reasoning, for all $d \geq 1$ we have ${\rm Aut}(({\mathbf Z}/(n))^d) = {\rm GL}_d(\mathbf Z/(n))$, where $d \times d$ mod $n$ matrices act on vectors in $(\mathbf Z/(n))^d$ in the natural way.
A very simple reason ${\rm Aut}(G \times G)$ is not pairs of automorphisms of $G$, when $G$ is a nontrivial group, is that there is always the swap automorphism $f(g,g') = (g',g)$.
A: No. Note that we have $\text{Aut}(\mathbb{Z_3}) \cong \mathbb{Z}_2$ so we can safely write
$$\text{Aut}(\mathbb{Z_3}) \times \text{Aut}(\mathbb{Z_3}) \cong \mathbb{Z}_2 \times \mathbb{Z}_2$$
However, if we consider an automorphism $\sigma$ in $\text{Aut}(\mathbb{Z_3} \times \mathbb{Z_3})$ with $\sigma(1,0) = (0,2)$ and $\sigma(0,1) = (1,0)$, we have $o(\sigma) = 4$ but there is no element of order $4$ in $\mathbb{Z}_2 \times \mathbb{Z}_2$.
