How to determine $\left|\operatorname{Aut}(\mathbb{Z}_2\times\mathbb{Z}_4)\right|$ I know that the group $\mathbb{Z}_2\times\mathbb{Z}_4$ has:
1 element of order 1 (AKA the identity)
3 elements of order 2
4 elements of order 4
I'm considering the set of all automorphisms on this group, denoted $\operatorname{Aut}(\mathbb{Z}_2\times\mathbb{Z}_4)$.
I know that:


*

*The automrphism needs to map the identity to the identity

*The automorphism needs to preserve the orders of the elements in the group.


So by my watch, the 3 elements of order 2 are permutated, and the 4 elements of order 4 are permuted.
I just wish to count how many automorphisms there are. I am wuite confused. Is it $3\times4=12$. I have a feeling this is wrong.

Note that $G$ is generated by two elements of order $4$. 
To count the possible automorphisms we can focus on the maps to these two generators. To the first generator we can map any of the $4$ elements of order $4$, giving $4$ possibilities. 
To the second generator we don't want to map the same element of order $4$ as the first generator. And we don't want to map the other element of order $4$ that shares the same subgroup of order $4$. That is, to the other generator we need to map one of the elements of order $4$ in the other subgroup of order $4$. This gives $2$ possibilities for the second generator, for a grand total of $4\times2=8$ automorphisms.
Note: You can also do this by noting that $G$ is generated by an element of order $4$ and an element of order $2$, but you have to be careful when assigning the element of order $2$. As I mentioned in the comments one element of order $2$ is in both of the subgroups of order $4$ and the other two are not. So an automorphism can't take that one one element of order $2$ to one of the other two. In particular the subgroup generated by the first generator of order $4$ contains that one element of order $2$, so it can't be the second generator.
