Isomorphism of two representations 
Let $\rho:G\to GL(V)$ and $\sigma:G\to GL(W)$ be two representations. We define the representation $\tau:G\to GL(\hom _k (V,W))$ by:


$$\tau_{g}\varphi=\sigma_{g}\circ\varphi\circ\rho_{g}^{-1}$$ for $g\in G$, $\varphi\in \hom{_{k}(V,W)}$. Let $V,W$ be finite dimensional. Prove that the representations $\tau$ and $\rho^*\otimes \sigma  $ are isomorphic (where $\rho^*$ is the dual representation of $\rho$, namely $\rho^*_g(\varphi)(v)=\varphi (\rho_g^{-1}(v))$

My attempt:
Let us fix some basis for V, $\{e_{i}\}_{i=1}^n$.
We define $T:\hom_{k}(V,W)\to V^{*}\otimes W$ to be the natural isomorphism, namely:
$$T(\varphi)=\sum_{i=1}^{n}e_{i}'\otimes\varphi(e_{i})$$
with $\{e_i'\}$ the dual basis for $\{e_i\}$. Then we have:
$$(T\circ\tau_{g})\varphi=T(\sigma_{g}\circ\varphi\circ\rho_{g}^{-1})=\sum_{i=1}^{n}e_{i}'\otimes(\sigma_{g}\circ\varphi\circ\rho_{g}^{-1}(e_{i}))$$
On the one hand, and on the other hand:
$$ ((\rho^{*}\otimes\sigma)_{g}\circ T)(\varphi)=(\rho^{*}\otimes\sigma)_{g}(\sum_{i=1}^{n}e_{i}'\otimes\varphi(e_{i}))=\sum_{i=1}^{n}(\rho^{*}\otimes\sigma)_{g}(e_{i}'\otimes\varphi(e_{i}))=$$$$=\sum_{i=1}^{n}\rho_{g}^{*}(e_{i}')\otimes(\sigma_{g}\circ\varphi)(e_{i})$$
And I don't see why those two are equal. Maybe it's something I'm missing in the definition or a miscalculation.
Any help would be appreciated.
 A: Your $T$ is independent by the choice of the basis $\{e_i\}$ (to see better take the inverse map, that is defined in a canonical way without fixing a basis)
Thus for each $g\in G$ you can take $\{e_i\}$ basis of eigenvectors of $\rho_g$, so that
$\rho_g^*(e_i’)=\lambda_i’e_i’$
where $\lambda_i’=\frac{1}{\lambda}$ is the complex conjugate of $\lambda$ (if the group $G$ is finite, otherwise just the inverse $\frac{1}{\lambda}$). At this point you get
$$ ((\rho^{*}\otimes\sigma)_{g}\circ T)(\varphi)=(\rho^{*}\otimes\sigma)_{g}(\sum_{i=1}^{n}e_{i}'\otimes\varphi(e_{i}))=\sum_{i=1}^{n}(\rho^{*}\otimes\sigma)_{g}(e_{i}'\otimes\varphi(e_{i}))=$$$$=\sum_{i=1}^{n}\rho_{g}^{*}(e_{i}')\otimes(\sigma_{g}\circ\varphi)(e_{i})= \sum_{i=1}^{n}\lambda_i’e_{i}'\otimes(\sigma_{g}\circ\varphi)(e_{i}) = \sum_{i=1}^{n}e_{i}\otimes(\sigma_{g}\circ\varphi)(\lambda_i’e_{i})= \sum_{i=1}^{n}e_i’\otimes(\sigma_{g}\circ\varphi\circ \rho_{g^{-1}})(e_{i}) $$
You can do the same also without taking a basis of eignevectors, but you have to re-write $\rho_g^*(e_i’)$ with respect to the basis $\{e_i’\}$
