A statement I have seen often in the context of the application of representation theory to quantum mechanics to derive selection rules is the following:
Suppose $V$ carries a representation of a group $G$ and $\psi \in U \subseteq V$ is an element of some subrepresentation $U$ of $V$. Let $A \in O \subseteq \mathrm{End}(V)$ be an operator in a subrepresentation $O$ (where $\mathrm{End}(V) \cong V^\ast \otimes V$ carries the induced $G$-representation, i.e. given by conjugation). Then $A\psi$ lies in a subrepresentation isomorphic to $O \otimes U$. In other words, the space $OU$ is isomorphic as a representation to $O \otimes U$.
It's clear that $OU$ is a subrepresentation since if $A$ and $\psi$ are as before one has, for all $g \in G$
$g(A\psi) = (gAg^{-1})(g\psi) \in OU$
since $gAg^{-1} \in O$ and $g\psi \in U$. I don't quite see the stronger statement that $O U \cong O \otimes U$ is obviously true as is sometimes stated when discussing selection rules. I suppose one could show that the characters agree for both but I am not sure if that is the smartest way to go about it.
They are not in general isomorphic, but rather $OU$ lies in a subrepresentation isomorphic to $O \otimes U$. One can construct a $G$-linear map $f : O \otimes U \to OU$ by evaluation, i.e. by setting $f(A \otimes \psi) = A\psi$ and extending by linearity. $G$-linearity can be seen from
$$ gf(A \otimes \psi) = g(A\psi) = (gAg^{-1})(g\psi) = fg(A\psi). $$
$f$ is certainly surjective but may not be injective, since the dimension of $OU$ could be smaller than the dimension of the tensor product. For example if $U$ happens to be invariant under $O$ then $\dim{OU} \leq \dim{u} \leq \dim{U}\dim{O} = \dim{O \otimes U} $.