I'm trying to determine the order of the automorphism group of $\mathbb Z_p\times \mathbb Z_{p^2}$ for $p$ prime.
What I've done: Let $(1,0)\mapsto (x,y_0)$ and $(0,1)\mapsto(z,t)$. For this map to be a homomorphism, we must have $(p,0)\mapsto (0,0)$ and $(0,p^2)\mapsto (0,0)$, so $y_0$ must be divisible by $p$; write $y_0=py$. So any homomorphism is given by $(0,1)\mapsto (x,py)$ and $(0,1)\mapsto (z,t)$ with $0\le x,y,z \le p-1,\ 0\le t\le p^2-1 $. (By the way, is this argument of determining whether a map from $\mathbb Z_p\times \mathbb Z_{p^2}$ to itself a group homomorphism complete?)
Now an automorphism is an invertible homomorphism. Clearly, if $t$ is divisible by $p$, then the image consists of elements of the form $(r,2s)$ and thus the homomorphism is not surjective. So $p$ must not divide $t$. Also, I have the hypothesis that $p$ must not divide $x$, but I don't see why this should be the case in general.
I tried to prove that $p \nmid x$ by converse (this worked in the case $p \nmid t)$: we have $(1,0)\mapsto (px,py),\ (0,1)\mapsto (z,t)$, so $(r,s)\mapsto (rpx+sz, rpy+st)$, but what's bad about this?
Also, I don't know whether the above two conditions are sufficient for the corresponding map to be invertible, and if so, I have no idea how one proves this.