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
$$V \otimes W = (V_0 \otimes W_0) \oplus (V_0 \otimes W^H) \oplus (V^H \otimes W_0) \oplus (V^H \otimes W^H)$$
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*}