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?