How can I compute $\mathrm{Aut}(\mathbb{Z}/4\mathbb{Z}\times\mathbb{Z}/6\mathbb{Z})$? How can I compute $\mathrm{Aut}(\mathbb{Z}/4\mathbb{Z}\times\mathbb{Z}/6\mathbb{Z})$?
If $\varphi$ is one of this automorphism then $\varphi((1,0))=(1,0),(1,3),(3,3),(3,0)$ and $\varphi((0,1))=(0,1),(0,5),(2,1),(2,5)$.
So it seems to me that this groups contains 16 elements, am i right? And if I'm right there is a way to understand to which group of cardinality 16 this is isomorphic to?
EDIT: $\mathbb{Z}/4\mathbb{Z}\times\mathbb{Z}/6\mathbb{Z}\cong\mathbb{Z}/8\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}$. And $\mathbb{Z}/8\mathbb{Z}$ and $\mathbb{Z}/3\mathbb{Z}$ are characteristic (am I right?), so $\mathrm{Aut}(\mathbb{Z}/8\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z})\cong\mathrm{Aut}(\mathbb{Z}/8\mathbb{Z})\times\mathrm{Aut}(\mathbb{Z}/3\mathbb{Z})\cong\mathbb{Z}/8\mathbb{Z}^*\times\mathbb{Z}/3\mathbb{Z}^*$ and this group has cardinality 8, right? Where is my mistake?
 A: $G = \mathbb{Z}_4\times \mathbb{Z}_6 = \mathbb{Z}_4\times (\mathbb{Z}_2\times \mathbb{Z}_3)$ where $\mathbb{Z}_4\times \mathbb{Z}_2$ and $\mathbb{Z}_3$ are characteristic in $G$. It's not hard to show that the automorphism group of a group that is the direct product of two characteristic subgroups is the direct product of the automorphism groups of the two characteristic subgroups.
As $\mathop{Aut}(\mathbb{Z}_3)$ is easily dealt with (it gives you the factor $\mathbb{Z}_2$) you are left with determining $\mathop{Aut}(\mathbb{Z}_4\times \mathbb{Z}_2)$.
$\mathbb{Z}_4\times \mathbb{Z}_2$ has four elements of order 4 (there are further 3 elements of order 2 and the neutral element). An automorphism of $\mathbb{Z}_4\times \mathbb{Z}_2$ permutes the four elements of order 4, but maps the inverse of each element of order $4$ on the inverse of its image [This trivial remark explains why the automorphism group is not the full symmetric group $S_4$.]. Even if only one element of order 2 is the square of a(ll) element(s) of order 4, the image of the elements of order 2 are determined by the images of the elements of order 4.
I guess you can finish off now.
PS: If you like geometry, label the four corners of a square with the four elements of order 4 where the opposing corner is labeled with the inverse. [Caution: m.k.'s $D_8$ is the dihedral group with $8$ elements, mathematicians not doing finite group theory call this group $D_4$.] 
