The question is phrased exactly as in the title.
I've tried listing the elements of this group, but there are just too many of them and I don't think that it would be reasonable to work with multiplication tables here because of the sheer order of the group.
I'm not interested very much in the actual answer to this question but more so in the way of thinking that one could employ to tackle problems like this. What should I do first? Should I try to think of a suitable $1-1$ homomorphism and then maybe reduce the target space until it becomes onto as well...
I'm really stuck so any and all help would be much appreciated. Thank you in advance!
We are looking at the automorphisms of the group $A:=\mathbb{Z}_6\oplus\mathbb{Z}_6=(\mathbb{Z}_2\oplus\mathbb{Z}_2)\oplus(\mathbb{Z}_3\oplus\mathbb{Z}_3)$.
Clearly, as the orders of elements are preserved, any automorphism of $A$ keeps the $2$- and $3$- parts invariant. So the automorphism group of $A$ is a subgroup of $\text{GL}(2,2)\times\text{GL}(2,3)$; and conversely every element of this product gives an automorphism of $A$.
So the answer to the part you are not very interested in is $\text{GL}(2,2)\times\text{GL}(2,3)$.