My attempt was basically to find all the subgroups of $G=\mathbb{Z}_3 \times \mathbb{Z}_9$ and all the automorphisms of $G$, and to try by hand. This is what I've done for the groups:
$G$ has (except for the trivial subgroups $\{e\}$ and $G$) subgroups of order 3 and 9. The subgroups of order 3 are cyclic, and so they are generated by an element of order 3. The number of these subgroups is $8/ \phi(3)=4$ (where 8 is the number of elements of order 3 in $G$), and they are $$\langle(1,0)\rangle,\ \langle(0,3)\rangle,\ \langle(1,3)\rangle,\ \langle(1,6)\rangle$$ The subgroups of order 9 are cyclic or isomorphic to $\mathbb{Z}_3 \times \mathbb{Z}_3$. In the first case we do the same thing: the number of elements of order 9 is $3\cdot \phi(9)=18$ and the number of cyclic subgroups of order 9 is $18/\phi(9)=3$, and they are $$\langle(0,1)\rangle,\ \langle(1,1)\rangle,\ \langle(2,1)\rangle$$ In the end, there is only one subgroup isomorphic to $\mathbb{Z}_3 \times \mathbb{Z}_3$ that is $\mathbb{Z}_3 \times \langle 3 \rangle$ because it contains all the elements of order 3, that are 8.
At this point, my doubts are:
- I don't know how to determine $\text{Aut}(G)$
- I really want to know if there is a faster way to do this excersise
