Semidirect Products of the form $(\mathbb{Z}_p \times \mathbb{Z}_p)\rtimes_{\phi}\operatorname{GL}(2,p)$ What are the different (non-isomorphic) semidirect products $(\mathbb{Z}_p \times \mathbb{Z}_p)\rtimes_{\phi}\operatorname{GL}(2,p)$, when $\phi \colon \operatorname{GL}(2,p)\rightarrow\operatorname{Aut}(\mathbb{Z}_p \times \mathbb{Z}_p)$ is injective homomorphism? ($p$ is any prime).
 A: The automorphism group of $\mathbb{Z}_p\times\mathbb{Z}_p$ is isomorphic to $\mathrm{GL}(2,p)$ (view $\mathbb{Z}_p\times\mathbb{Z}_p$ as a two-dimensional vector space over $\mathbb{F}_p$, the field of $p$ elements). So your $\phi$ is just an automorphism of $\mathrm{Aut}(\mathbb{Z}_p\times\mathbb{Z}_p)$.
If $\phi$ is the identity, then you have the holomorph of $\mathbb{Z}_p\times\mathbb{Z}_p$, where the holomorph of $G$ is $\mathrm{Hol}(G) = G\rtimes \mathrm{Aut}(G)$, with the obvious action.
And so the question comes down to the following: if $K\rtimes_{\phi} H$ is a semidirect product, and $\psi\colon H\to H$ is an automorphism, is $K\rtimes_{\phi\circ\psi}H$ isomorphic to $K\rtimes_\phi H$?
The answer is yes. Define $F\colon K\rtimes_{\phi}H\to K\rtimes_{\phi\circ \psi}H$ by $F(h,k) = (\psi^{-1}(h),k)$. This is trivially a bijection, so we just need to show that it is a group homomorphism.
If $(h,k),(y,x)\in K\rtimes_{\phi}H$, we have:
$$\begin{align*}
(h,k)(y,x) &= (hy, k^{\phi(y)}x).\\
F\bigl( (h,k)(y,x)\bigr) &= (\psi^{-1}(hy),k^{\phi(y)}x).\\
F(h,k)F(y,x) &= (\psi^{-1}(h),k)(\psi^{-1}(y),x)\\
&= \bigl( \psi^{-1}(h)\psi^{-1}(y), k^{\phi\circ\psi(\psi^{-1}(y))}x\bigr)\\
&= \bigl(\psi^{-1}(hy), k^{\phi(y)}x\bigr)\\
&= F\bigl( (h,k)(y,x)\bigr).
\end{align*}$$
So $F$ is a group homomorphism, hence an isomorphism.
Applying this to the particular case you have, there is exactly one such semidirect product, isomorphic to the holomorph of $\mathbb{Z}_p\times\mathbb{Z}_p$. 
