= Solution
The character of $(\rho,V)$ is
$$
\chi_V(g)=\operatorname{tr}\rho(g).
$$
The stated action is a representation because successive action by $(g_2,h_2)$ and $(g_1,h_1)$ gives
$$
\sigma(h_1h_2)\,\alpha\,\rho(g_2^{-1}g_1^{-1})
=\sigma(h_1h_2)\,\alpha\,\rho((g_1g_2)^{-1}).
$$
Identifying $\operatorname{Hom}_{\mathbb C}(V,W)$ with $V^*\otimes W$ gives the <character of a Hom representation>
$$
\boxed{\chi_{\operatorname{Hom}(V,W)}(g,h)
=\chi_V(g^{-1})\chi_W(h)
=\overline{\chi_V(g)}\,\chi_W(h)}.
$$
Here the final equality follows after unitarizing the finite-group representation.
The permutation character of $\mathbb CG$ counts fixed points:
$$
\chi_{\mathbb CG}(g,h)
=\#\{x\in G:gxh^{-1}=x\}
=\#\{x:x^{-1}gx=h\}.
$$
Thus
$$
\chi_{\mathbb CG}(g,h)=
\begin{cases}
|C_G(g)|,&g\text{ and }h\text{ are conjugate},\\
0,&\text{otherwise}.
\end{cases}
$$
On the other hand, the character of
$\bigoplus_i\operatorname{Hom}(V_i,V_i)$ is
$$
\sum_i\overline{\chi_i(g)}\chi_i(h).
$$
The column form of <character orthogonality> says that this has exactly the same two values above. Complex representations of a finite <group> are semisimple, and semisimple representations with the same character are isomorphic. Hence the <two-sided regular representation decomposition> is
$$
\boxed{\mathbb CG\cong
\bigoplus_{i=1}^r\operatorname{Hom}_{\mathbb C}(V_i,V_i)}
$$
as a $G\times G$ representation.
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