Suppose that we have $G=\mathbb{Z}_p\times \mathbb{Z}_p$, where $p\in \mathbb{P}$. I know that every arbitrary permutation of the elements of $G$ such that $ \forall x \in G: {\rm ord}(f(x))=o(x) \Rightarrow f\in {\rm Aut}(G)$. Therefore, we can write $G$ group as follows:$$G=\{x^i,y^j,x^ky^l\mid i,j,k,l\in \{0,...,p-1\}\}=\langle x,y\rangle .$$ where $x=(1,0)$ and $y=(0,1)$. I know that $f(e)=e$ and $(\forall x\in G) \land (\forall f\in{\rm Aut}(G)):{\rm ord}(f(x))={\rm ord}(x)$ But, $\forall x\in G:{\rm ord}(x)\mid|\mathbb{Z}_p|\Rightarrow ({\rm ord}(x)=1) \lor ({\rm ord}(x)=p)$. Therefore,$\forall x \in G: ({\rm ord}(x)=p) \lor (x=e)$.
If I can define the correspondent elements of $f(x)=\bar x$ and $f(y)=\bar y$ ($\bar x,\bar y \in G$). I have $p^2-1$ candidate elements for $\bar x$ and $p^2-p$ candidate elements for $\bar y$ (because $ord(\bar x)=p,\forall \bar x \in G $ and $f(\bar y)\notin \langle\bar x\rangle$. Thus,$|{\rm Aut}(G)|=(p^2-1)(p^2-p)$.
Is it correct? If it's not, can you please indicate my mistakes? Thank you very much in advance!!
To expand on Greg's comment, your answer is indeed correct. An automorphism of $\mathbb{Z}_p\times\mathbb{Z}_p$ can be thought of as an invertible 2×2 matrix with coefficients in $\mathbb{Z}_p$. Thus $|\text{Aut}(\mathbb{Z}_p\times\mathbb{Z}_p)|=|\text{GL}_2(\mathbb{Z}_p)|$. So how many of these matrices are there? Giving a 2×2 invertible matrix is the same as finding two linearly independent vectors over $\mathbb{Z}_p$. Well, let's take any nonzero vector $v$. There are $p^2-1$ such choices. Now we need to see how many vectors are linearly independent from $v$. But it's actually much easier to see how many linearly dependent vectors there are to $v$. Such vectors must of course be of the form $av$ for some $a\in \mathbb{Z}_p$. So there are $p$ vectors linearly dependent to $v$, and hence $p^2-p$ vectors linearly independent to $v$. Putting it all together, there are $(p^2-1)(p^2-p)$ pairs of linearly independent vectors.