What is the order of the automorphism group of the finite group $G=\mathbb{Z}/5\mathbb{Z}\oplus \mathbb{Z}/25\mathbb{Z}.$ 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?
 A: Let $G=Z_p\times Z_{p^2}$ where $p$ is prime and $Z_n$ is cyclic of order $n$.
Then $G$ has $p^3-p^2$ elements of order $p^2$. Let $a$ be any of them. Then $G$
has $p^2-p$ elements of order $p$ outside $\left<a\right>$, call one of them $b$.
There is a unique automorphism sending the standard generators of $G$ to $a$ and $b$,.
Therefore $G$ has
$$(p^3-p^2)(p^2-p)=p^3(p-1)^2$$
automorphisms.
A: 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.
