Would it be correct to prove this statement as follows?
$(0,0)$ must always be sent to $(0,0)$ under any $\phi\in$ Aut($\mathbb{Z}_2\times \mathbb{Z}_2)$. Now, for Out($\mathbb{Z}_2\times \mathbb{Z}_2)$, the only possible permutations are on the subset $X=\{(1,0), (0,1), (1,1)\}$. Since $|S_X|=3!=6$, $|$Aut($\mathbb{Z}_2\times \mathbb{Z}_2)|=6$. Now, for any $\phi \in$ Aut($\mathbb{Z}_2\times \mathbb{Z}_2)$, $\phi_1\phi_2[(a,b)]=\phi_2\phi_1[(a,b)]$, thus Aut($\mathbb{Z}_2\times \mathbb{Z}_2)$ is abelian. We can conclude, therefore, that Aut($\mathbb{Z}_2\times \mathbb{Z}_2)\cong \mathbb{Z}_6$.
I'm somewhat concerned about the level of rigour about the fact that Aut($\mathbb{Z}_2\times \mathbb{Z}_2)$ is abelian. We can check every element, but if we view the elements of Aut$(\mathbb{Z}_2\times \mathbb{Z}_2)$ as transpositions of permutations, we know that they do commute.