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Let $\lambda$ be a partition of $n$, and let $T$ be the standard tableau associated to $\lambda$ (write the Young diagram of $\lambda$ down and fill in the boxes with $1$ through $n$ left to right, from top row to bottom). Let $T'$ and $\lambda'$ denote the conjugates (i.e., flips about the diagonal) of $T$ and $\lambda$, respectively. Let's say $\lambda = \lambda'$.

Let $R(T)$ and $C(T) = R(T')$ be the row-stabilizer and column-stabilizer of $T$ (i.e., the subgroup of $S_n$ that stabilizes the numbers in the rows (resp., columns) of $T$ set-theoretically --- e.g., $R(T) = S_{\lambda_1}\times S_{\lambda_2}\cdots$.)

Let $a_T := \sum_{g\in R(T)} g$, and similarly for $T'$. (The notation arises from the usual definition of Young symmetrizers, which project the group algebra to a copy of the irreducible representation corresponding to $\lambda$.)

When is the action of $a_T a_{T'}$ on the irreducible representation corresponding to $\lambda$ nonzero?

I really only care about the case of $\lambda = (k,\ldots,k)$ [$k$ times], a $k\times k$ square. Note that $a_T$ projects onto the invariants under $R(T)$ of the irrep. corresponding to $\lambda$, which is one-dimensional by Frobenius reciprocity/knowledge of the decomposition of $\mathrm{Ind}_{R(T)}^{S_n}(\mathrm{triv})$ (similarly for $a_{T'}$).

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