Bipartite Version of Schur's Lemma Let $V$ be a finite-dimensional vector space, let $\pi$ be a representation of some group $G$ (I'm really interested in $G=SO(n)$) on $V$, and suppose that a linear operator $S$ on $V\otimes V$ commutes with all $U\otimes U$, where $U=\pi(g)$ for some $g\in G$. I think it should be the case that $S$ is a linear combination of the identity and the linear operator $\text{SWAP}:x\otimes y\mapsto y\otimes x$. Is this true? I'm really not sure where to start.
Edit: As Qiaochu points out, $\pi$ should of course be an irrep.
 A: This is true for $G = GL(V)$ acting on $V$ (and the ground field doesn't hae characteristic $2$), and this is a special case of Schur-Weyl duality. If we're working over $\mathbb{C}$ we can also take $G = U(n)$ acting on $\mathbb{C}^n$. There's no reason it should be true in general; note that you didn't make any kind of irreducibility assumption on $V$, so $G$ could be acting trivially.
What we can say in general is that $V \otimes V$ always decomposes under the action of the flip map as a direct sum
$$V \otimes V \cong \text{Sym}^2(V) \oplus \text{Alt}^2(V)$$
of the symmetric and the antisymmetric tensors, where the symmetric tensors are the $+1$ eigenspace and the antisymmetric tensors are the $-1$ eigenspace. (This is why the ground field can't have characteristic $2$.) If both of these are irreducible representations of $G$ then Schur's lemma applied to this decomposition implies the desired result (and the converse holds if the action of $G$ is completely reducible, e.g. if $G$ is finite or compact). This requires in particular that $V$ is an irreducible representation of $G$ but this condition is only necessary, not sufficient.
The desired result is not true for $G = O(n)$ acting on $\mathbb{R}^n$ (which is irreducible). The issue is that, by definition, $G$ leaves $\sum_{i=1}^n x_i^2$ is invariant, which spans a trivial subrepresentation of $\text{Sym}^2(\mathbb{R}^n)$. So the symmetric square is reducible. The exterior square is also reducible when $n = 4$; in this case $\text{Alt}^2(\mathbb{R}^4)$ is famously known to break up into two further irreducible representations due to the action of the Hodge star operator. $\text{Alt}^2(\mathbb{R}^n)$ is an irreducible representation of $O(n)$ for $n = 2, 3$ and $n \ge 5$, though.
