Is there an isomorphism $\text{Hom}_R(R \otimes_k V, R \otimes_k W) \cong R \otimes_k W \otimes_k V^*$? Suppose that $R = k[x_1, \dots, x_n]$, $G$ is a finite group acting on $R$, and $V$ and $W$ are finite dimensional $G$-modules (perhaps one-dimensional?). 

Is there an isomorphism (of $kG$-modules?)
  $$
\text{Hom}_R(R \otimes_k V, R \otimes_k W) \cong R \otimes_k W \otimes_k V^*?
$$

Here I'm viewing $R \otimes_k V$ and $R \otimes_k W$ as left $R$-modules in the usual ring-theoretic way. Whenever $kG$ acts on a tensor product, the action splits over the tensor. Unless I'm getting confused, $\text{Hom}_R(R \otimes_k V, R \otimes_k W)$ is a left $kG$-module via
$$
(g \cdot \phi)(r \otimes v) = g \cdot \phi(g^{-1} \cdot(r \otimes v)) = g \cdot \phi((g^{-1} \cdot r) \otimes (g^{-1} \cdot v)).
$$
I tried using hom-tensor adjunction, but it's not clear to me if that holds here (tensoring $kG$-modules seems slightly different to tensoring modules over rings).
 A: You are right in saying that tensoring $k[G]$-modules like you do here is different than tensoring modules over rings, because in general if $A$ is a ring and $M$ and $N$ are $A$-modules, $M\otimes N$ is not an $A$-module, but simply an $(A\otimes A)$-module. The miracle here is that $k[G]$ is much more than a ring, it is a Hopf algebra, and in particular it has a comultiplication $k[G]\to k[G]\otimes k[G]$ (given by $g\mapsto g\otimes g$) which allows us to turn any $(k[G]\otimes k[G])$-module into a $k[G]$-module.
Now about your question. The first step is to see that at least the isomorphism holds as $k$-vector spaces, which I think you know how to do:
$$ \operatorname{Hom}_R(R\otimes_k V, R\otimes_k W)\simeq \operatorname{Hom}_k(V, R\otimes_k W)\simeq R\otimes_k W\otimes_k V^* $$
with the isomorphism $\Phi: R\otimes_k W\otimes_k V^*\to \operatorname{Hom}_R(R\otimes_k V, R\otimes_k W)$ given by
$$ r\otimes w\otimes f \mapsto \left( a\otimes v\mapsto f(v)\cdot ar\otimes w\right).$$
The fact that it is an isomorphism hinges on the fact that $V$ is finite-dimensional.
So we just have to check that this respects the $G$-actions. Now:
$$\Phi(gr\otimes gw\otimes gf)(a\otimes v) = f(g^{-1}v)\cdot a(gr)\otimes (gw)$$
and 
$$\begin{align*}
g(\Phi(r\otimes w\otimes f))(a\otimes v) &= g\left( \Phi(r\otimes w\otimes f)(g^{-1}a\otimes g^{-1}v) \right)\\
&= g\left( f(g^{-1}v)\cdot (g^{-1}a)r\otimes w \right) \\
&= f(g^{-1}v)\cdot a(gr)\otimes gw.
\end{align*}$$
So indeed this is an isomorphism of $k[G]$-modules. Note that this requires that the action of $G$ on $R$ is by algebra automorphisms, which I think is implicit in your question (so we should have $g(ab)=(ga)(gb)$ if $a,b\in R$).
