Automorphism group of $\mathbb{Z}_p\times \mathbb{Z}_p$

An automorphism of a vector space is defined by its action on a basis. So, take the unique automorphism determined by

$(1,0)\rightarrow (0,1)$ and $(0,1)\rightarrow (1,0)$.

As suggested by vadim123, the automorphism group is isomorphic to $\operatorname{GL}(2, p)$, the group of $2 \times 2$ invertible matrices over $\Bbb{Z}$.

If $p = 2$, the group $\operatorname{GL}(2, 2)$ is isomorphic to $S_{3}$, so you should be able to determine the involutions, which are $$ \begin{bmatrix}0&1\\1&0\end{bmatrix}, \quad \begin{bmatrix}1&1\\0&1\end{bmatrix}, \quad \begin{bmatrix}1&0\\1&1\end{bmatrix}. $$

If $p > 2$, you will have $$ c = \begin{bmatrix}-1&0\\0&-1\end{bmatrix}, $$ a central element, and then the conjugacy class of $$ b = \begin{bmatrix}1&0\\0&-1\end{bmatrix}, $$ which has order $$ \frac{\lvert \operatorname{GL}(2, p) \rvert}{\lvert C_{\operatorname{GL}(2, p)}(b) \rvert} = \frac{(p^{2} - 1)(p^{2} - p)}{(p-1)^{2}} = (p+1)p. $$