Let $(\rho,V)$ be a representation of $G$, and we consider the following action of $G\times G$ on $\hom(V,V)$:$$\forall(g,h)\in G\times G:(g,h).\phi=\rho_g\circ\phi\circ\rho_{h^-1}$$ Find an isomorphism between $\hom(V,V)^{G\times G}$ and $\hom(V^G,V^G)$.
First, clarification of definitions:$$\hom(V,V)^{G\times G}=\{\phi\in \hom (V,V):\forall(g,h)\in G\times G, \phi=\rho_g\circ\phi\circ\rho_{h^-1} \}$$ and $$V^G=\{v\in V:\forall g\in G,\rho_g (v)=v \}$$ Where $\rho_g$ is my notation to the linear transformation corresponding to $g$.
My attempt: I first showed that if $v\in V$, considering the pair $(g,e)\in G\times G$, I have: $$(\rho_g\circ\phi\circ\rho_e)(v)=\phi(v)$$ and therefore $$\rho_g(\phi(v))=\phi(v)$$ So $\phi(V)\subset V^G$. Now I can define $\psi:\hom(V,V)^{G\times G}\to\hom(V^G)$ by $\psi(\phi)=\phi|_{V^G}$, but I can't see if this is one-to-one\onto. I feel like there's a trick I'm missing.
Any hint would be appreciated.