Group action on vector space Let $G$ be a group acting on (complex) vector spaces $V, W$ and let $G$ act trivially on the vector space $U$.
Let $Hom^G(V, W)$ denote the linear transformations $T:V \to W$ that respect the group action, that is, $g \cdot T(v) = T(g \cdot v)$ for all $v \in V$.
$W \otimes U$ has a natural $G$-action given by $g \cdot (w \otimes u) = gw \otimes gu = gw \otimes u$.

How do I see $Hom^G(V, W \otimes U) \cong Hom^G(V, W) \otimes U$?

There is a natural map going backwards given by sending $(T: V \to W) \otimes u$ to the map $v \mapsto T(v) \otimes u$. 
This second map respects the $G$-action because $T$ respects the $G$-action, and $U$ has trivial $G$-action.

I don't believe this is true in full generality; one needs $U$ to be finite-dimensional.
In the case where $G$ is the trivial group, this amounts to
$$\text{Hom}(V,W\otimes U)\cong\text{Hom}(V,W)\otimes U.$$
Specialising further to $W=\Bbb C$ (one-dimensional) we get
$$\text{Hom}(V,U)\cong\text{Hom}(V,\Bbb C)\otimes U.\tag{*}$$
The natural map from the right of (*) to the left has its image
the homomorphisms from $V$ to $U$ with finite-dimensional image,
so is not surjective. 
But if $U$ is finite-dimensional one can write $U$ as finite direct
sum of trivial one-dimensional modules. As both functors
$U\mapsto \text{Hom}^G(V,W\otimes U)$ and $U\mapsto \text{Hom}^G(V,W)\otimes U$ preserve finite direct sums, we reduce to the case $U=\Bbb C$ with trivial $G$-action, in which case the result is immediate. 
