Consider the vector space $\mathbb C[G]$ of functions $f: G \longrightarrow \mathbb{C}$ where $G$ is a finite group, or equivalently a vector space of all formal linear combinations of elements of $G$ over the complex numbers. How can I prove the map $$\phi : \mathbb{C} [G] \longrightarrow \prod_{\rho} \text{End}(V_{\rho})$$ $$f \mapsto (\rho(f) )_{\rho}$$ where $\rho(f) = \sum_{g \in G} f(g) \rho(g)$ (where $\rho$ runs over all the isomorphism classes of irreducible representations of $G$) is both a ring isomorphism and also an intertwining isomorphism of $G \times G$ where our action, given $T \in \text{End}(V_{\rho})$, is $(g,h)\cdot T := \rho(g) \circ T \circ \rho(h)^{-1}$? We define ring multiplication by $(f_1 \ast f_2)(g) = \sum_{g = x y} f_1(x) f_2(y)$ in $\mathbb{C}[G]$. Does this isomorphism lead to a more natural decomposition of the regular representation $\mathbb{C}[G]$?
I know how to show that each map $\rho : G \longrightarrow \text{GL}(V_{\rho})$ extends to $\phi$ by linearity. I can also see how ring multiplication is respected by $\phi$. I also know that we can show the map is injective because $\dim \mathbb{C}[G] = |G|$, but also $$\dim (\prod_{\rho} \text{End}(V_{\rho}))=\sum_\rho \dim (\text{End}(V_\rho))=\sum_\rho \dim(V^{\ast}_\rho \otimes V_\rho) =$$ $$\sum_\rho \dim(V^{\ast}_\rho) \dim(V_\rho) =\sum_\rho \dim(V_\rho)^2 = |G|$$ by the theorem that the sum of the squares of the degrees of the irreducible representations is equal to the order of the group.
I can't seem to work out how as a linear transformation it intertwines with the action on $G \times G$ described above or how the convolution multiplication defined above is respected by $\phi$ as a ring isomorphism.