If we have two different homomorphisms $\phi_1,\phi_2:\mathbb{Z}/2\mathbb{Z} \to \text{Aut}(\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z})=\text{GL}_2(\mathbb{Z}/p\mathbb{Z})$ they will give the same group $\mathbb{Z}/2\mathbb{Z}\rtimes_\phi \mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}$ if they are related by conjugation (Let $\mathbb{Z}/2\mathbb{Z}=\{0,1\}$, $\phi_1:0 \mapsto \left(\begin{matrix}1 0 \\ 0 1 \end{matrix} \right), 1 \mapsto \left(\begin{matrix}-1 0 \\ 0 1 \end{matrix} \right) $ and $\phi_1:0 \mapsto \left(\begin{matrix}1 0 \\ 0 1 \end{matrix} \right), 1 \mapsto g\left(\begin{matrix}-1 0 \\ 0 1 \end{matrix} \right)g^{-1} $ for some $g \in \text{GL}_2(\mathbb{Z}/p\mathbb{Z})$).
I searched online and found this property without proof: Suppose we have two groups $G$,$H$. If $\phi:H\to \text{Aut}(G)$ and $f \in \text{Aut}(G)$, show that $$G \rtimes_\phi H \cong G \rtimes_{f\phi}H$$
Can someone provide the proof of this statement?
I’ve figured out myself somehow, we have this commutative diagram https://i.stack.imgur.com/wduGv.jpg
And therefore they’re isomorphic