I haven't found anywhere that discusses representations of the symmetric group in a way that applies to this situation, and I'd appreciate any pointers/resources/phrases-to-Google.
Consider a vector space $V$, and form the $N$-fold tensor product $\otimes^NV$. The symmetric group $S_N$ has a natural action on $\otimes^N V$ given by $\sigma\cdot( v_1\otimes\cdots\otimes v_N)=v_{\sigma(1)}\otimes\cdots\otimes v_{\sigma(N)}$. I want to consider the set of all linear operators $L$ on $\otimes^N V$ satisfying $\sigma L=L\sigma$. I'll call this set $\mathcal G$. It clearly forms a semigroup (and if it makes things easier, I'm happy to restrict $L$ to be invertible, or unitary, so that $\mathcal G$ forms a group).
The action of $\mathcal G$ on $\otimes ^N V$ is not irreducible. Two obvious invariant subspaces are the symmetric subspace, which has dimension $\left(\begin{smallmatrix}N+\text{dim}(V)-1\\N\end{smallmatrix}\right)$, and the antisymmetric subspace, which has dimension $\left(\begin{smallmatrix}\text{dim}(V)\\N\end{smallmatrix}\right)$. Each of these is irreducible.
My question is How many irreducible subspaces are there? For $N>2$, there clearly is at least one more irreducible subspace, since the dimensions of the symmetric and antisymmetric subspaces don't add up to $\text{dim}(V)^N$. But is this third subspace irreducible, or does it decompose into some number of irreducible subspaces?
You can read more context for this question here, where it's phrased in the language of multi-particle quantum mechanics.