$(V \otimes W)^H = V^H \otimes W^H$? Let $G$ be a group, $H$ a subgroup, and $(\pi_1,V),(\pi_2,W)$ finite dimensional complex representations of $G$.  Let $V^H$ and $W^H$ the spaces of $H$-fixed vectors in $V$ and $W$.  Let $(\pi_1 \otimes \pi_2,V \otimes W)$ be the tensor product representation of $G$.  Do we have
$$(V \otimes W)^H = V^H \otimes W^H?$$
If $G$ is a Hausdorff topological group, $H$ a compact subgroup, and $\pi_1, \pi_2$ are continuous, then I think the answer is yes.  In that case, the $H$-stable subspaces $V^H$ and $W^H$ have $H$-invariant complements, say $V_0$ and $W_0$, and then the claim follows from looking at the direct sum decomposition
$$V \otimes W = (V_0 \otimes W_0) \oplus (V_0 \otimes W^H) \oplus (V^H \otimes W_0) \oplus (V^H \otimes W^H)$$
 A: Consider the group $G = H = \mathbb{Z}/2$, which acts on $V = W = \mathbb{C}^2$ by swapping the coordinates.
Then $V^H = W^H = \{(x,x) \mid x \in \mathbb{C}\} = \langle (1,1) \rangle$ and thus
$$
    V^H \otimes W^H
  = \langle (1,1) \rangle \otimes \langle (1,1) \rangle
  = \langle (1,1) \otimes (1,1) \rangle.
$$
Then the vector $e_1 \otimes e_2 + e_2 \otimes e_1$ is $H$-invariant but not contained in $V^H \otimes W^H$, which shows that $(V \otimes W)^H \supsetneq V^H \otimes W^H$.

PS:
I think it is nice to look at the case $H = \mathbb{Z}/2$ in a bit more detail:
If the linear map $f \colon V \to V$ is given by the action of $\overline{1} \in H$ on $V$, i.e. $f(v) = \overline{1} \cdot v$ for all $v \in V$, then $f^2 = \operatorname{id}_V$.
It then follows that $f$ is diagonalizable with possible eigenvalues $1$ and $-1$, so we have that
$$
    V
  = V_+ \oplus V_-
$$
where $V_+$ is the eigenspace of $f$ for the eigenvalue $1$, and $V_-$ the  eigenspace for the eigenvalue $-1$.
Note that this is a decomposition into $H$-subrepresentations, and that $V_+ = V^H$.
Similarly we have a decomposition $W = W_+ \oplus W_-$.
It then follows that
$$
    V \otimes W
  =        (V_+ \otimes W_+)
    \oplus (V_+ \otimes W_-)
    \oplus (V_- \otimes W_+)
    \oplus (V_- \otimes W_-).
$$
This is a decomposition into subrepresentations with $V^H \otimes W^H = V_+ \otimes W_+$ being one of the summands.
But we can see that $H$ also acts trivially on the summand $V_- \otimes W_-$ because
$$
    \overline{1} \cdot (v \otimes w)
  = (\overline{1} \cdot v) \otimes (\overline{1} \cdot w)
  = (-v) \otimes (-w)
  = v \otimes w
$$
for every simple tensor $v \otimes w \in V_- \otimes W_-$.
In a similar way we see that $\overline{1}$ acts by multiplication with $-1$ on the two summands $V_+ \otimes W_-$ and $V_- \otimes W_+$, so we see alltogether that
$$
    (V \otimes W)^H
  = (V_+ \otimes W_+) \oplus (V_- \otimes W_-)
  = (V^H \otimes W^H) \oplus (V_- \otimes W_-).
$$

In the above example we have that
$$
    V_+
  = W_+
  = V^H
  = W^H
  = \{ (x,x) \mid x \in \mathbb{C} \}
$$
and
$$
    V_-
  = W_-
  = \{ (x,-x) \mid x \in \mathbb{C} \}
$$
and therefore
\begin{align*}
   &\, (V \otimes W)^H \\
  =&\,        \langle (1,1) \otimes (1,1) \rangle
       \oplus \langle (1,-1) \otimes (1,-1) \rangle \\
  =&\, \langle (1,1) \otimes (1,1), (1,-1) \otimes (1,-1) \rangle \\
  =&\, \langle
            e_1 \otimes e_1
          + e_2 \otimes e_1
          + e_1 \otimes e_2
          + e_2 \otimes e_2,
            e_1 \otimes e_1
          - e_2 \otimes e_1
          - e_1 \otimes e_2
          + e_2 \otimes e_2
       \rangle \\
  =&\, \langle
         e_1 \otimes e_1 + e_2 \otimes e_2,
         e_1 \otimes e_2 + e_2 \otimes e_1
       \rangle
\end{align*}
