automorphism group of a group of order $p^2$, where $p$ is prime

Question

There is a corollary that states:

"If $|P|=p^2$ for some prime $p$, then $P$ is abelian. More precisely, $P$ is isomorphic to either $\mathbb{Z}_{p^2}$ or $\mathbb{Z}_p\times \mathbb{Z}_p$."

I know that the automorphism group of $\mathbb{Z}_p\times \mathbb{Z}_p$ is $GL_2(\mathbb{F}_p)$, which has order $p(p-1)^2(p+1)$, so |Aut($P$)|=$p(p-1)^2(p+1)$ if $P \cong \mathbb{Z}_p\times \mathbb{Z}_p$.

What if $P\cong \mathbb{Z}_{p^2}$? I know that then |Aut($P$)|$=p(p-1)$, but I don't know why. What is Aut($\mathbb{Z}_{p^2}$)?

Thanks.

Answer 1

$\phi \in $ Aut($\mathbb{Z}_{p^2}$) is uniquely determined by the image of a generator, which must be another generator.

The generators of $\mathbb{Z}_{p^2}$ are $\phi(p^2) = p(p-1)$ where $\phi$ is the Euler function, and so $$ |\text{Aut}(\mathbb{Z}_{p^2}) | = p(p-1)$$