What is the order of the automorphism group of the finite group $G=\mathbb{Z}/5\mathbb{Z}\oplus \mathbb{Z}/25\mathbb{Z}.$ Is the group $Aut(G)$ Abelian?
My attempt: If $\phi \in Aut(G)$, then $\phi$ sends generator to the generator. For this question, we need to count the total number of generator in $G$. There are $4$ generator in $\mathbb{Z}/5\mathbb{Z}$; and $20$ generator in $\mathbb{Z}/25\mathbb{Z}$. This implies $|Aut(G)|=80$.
I think $Aut(G)$ is not abelian, because a finite group $G$ has an abelian automorphism group iff $G$ is cyclic. In this case, $G$ is abelian, not cyclic.
Any one please suggest me whether this idea is correct for this question?
As you say there are 20 generators of $\mathbb{Z}/25\mathbb{Z}$. Thus there are 100 elements of order 25 in $G$. Therefore an automorphism of $G$ can map $(0,1)$ to 100 different places. Suppose it gets mapped to some element $x$. Now consider the subgroup $$\mathbb{Z}/5\mathbb{Z}\times\mathbb{Z}/5\mathbb{Z}\subseteq \mathbb{Z}/5\mathbb{Z}\times\mathbb{Z}/25\mathbb{Z}.$$
We know that $(1,0)$ has order $5$, so must map to an element $y$ in this subgroup. We know $5x$ is in this subgroup, generating a subgroup of size 5. In order for our map to be surjective, $(1,0)$ must map to one of the 20 other elements in $\mathbb{Z}/5\mathbb{Z}\times\mathbb{Z}/5$.
Thus there are $20\times 100=2000$ automorphisms.
If you believe this group is not commutative, then try writing down two elements which do not commute. Start simple.