rt.representation theory - Explicit computation of induced modules of semidirect products with the symmetric group

I suspect I know the answer, but I don't yet have a proof (not because I think it would be hard to prove, but because I didn't try really; when you see my guess, you'll likely want to believe it). The answer is stated not in the basis of simples, because I didn't compute the decomposition of $mathbb{C}[S_n/S_{pi}]$. However, it is stated in the tensor category S_n-mod, so that given that decomposition, you can easily adjust what I write here.

Fact: Let H be a finite dimensional semi-simple Hopf algebra (e.g. H=mathbb{C}[S_n]), and let $Vin H$-mod be an irrep. Let us regard H as an H-module via the left action. Then $Votimes Hcong H^{oplus dim(V)}$.

The proof of this fact is given as follows (see http://www-math.mit.edu/~etingof/tenscat.pdf, or Akhil's comments below):

$Hom_H(Votimes H,W)=Hom_H(H,^ast Votimes W) = widetilde{^ast Votimes W}$

On the other hand, $Hom_H(tilde{V}otimes H,W) = tilde{V}otimes Hom_H(H,W) = widetilde{Votimes W}$,

where $tilde{M}$ means we forget the module $M$ down to a vector space, which we use as a multiplicity space (just because the direct sum decomposition I asserted originally isn't canonically given, you just know that there's this multiplicity space)

(above we took right duals since I didn't assume $H$ is commutative or co-commutative; for $C[G]$ there is no need to distinguish.) One could (and should) be uncomfortable that we got duals on the one hand and not on the other. However, the standard representation for $S_n$ is special in that it is isomorphic to its own dual, by sending $e_i$ to $e^i$ (the point is that the standard rep for $S_n$ has a basis build into its definition).

The general fact above about Hopf algebras is used to relate Frobenius-Perron dimension for representations of Hopf algebras to ordinary dimension of the underlying vector space; indeed the regular representation is the unique eigenvector which realizes the Frobenius Perron dimension as an eigenvalue.

Okay so now we are considering $mathfrak{h}otimes mathbb{C}[S_n/S_{pi}]to mathbb{C}[S_n/S_{pi}]$. This is then isomorphic to $(mathfrak{h}otimes mathbb{C}[S_n])otimes_{S_pi}mathbf{1}$, where we tensor the trivial $S_pi$-module on the right. This is because $C[S_n]$ is a $S_n-S_pi$ bi-module, so that the map
$mathfrak{h}otimes mathbb{C}[S_n] to mathfrak{h}otimes mathbb{C}[S_n/S_pi]$ given by right multiplying with the symmetrizer $a_pi=sum_{gin S_pi} g$ is an $S_n$-morphism, and allows us to identify $mathfrak{h}otimes mathbb{C}[S_n]otimes_{S_pi}mathbf{1}$ with $mathfrak{h}otimes mathbb{C}[S_pi]$.
Morally, this is just because $S_pi$ acts on the right, while the other action is on the left.

[edited an error from preceding paragraph]

Together with the Fact, this implies that $mathfrak{h}otimes mathbb{C}[S_n/S_pi]$ is in fact just isomorphic to $mathbb{C}[S_n/S_pi]^{oplus dim(mathfrak{h})},$ which I should really write as
$mathbb{C}[S_n/S_pi]otimes tilde{mathfrak{h}^ast}$.

Well, now we have this function $c: mathfrak{h}to mathbb{C}$. We will project $mathbb{C}[S_n/S_pi]^{oplus dim(mathfrak{h})}$ (or rather $mathbb{C}[S_n/S_pi]otimes tilde{mathfrak{h}}$) to $mathbb{C}[S_n/S_pi]$ by just applying $c$ to the multiplicity space.

I haven't really proved that this last paragraph is what happens, but once one has applied "Fact" above, this seems like the only natural guess. I imagine verifying it would be pretty straightforward.

Note that it doesn't seem to matter how $mathbb{C}[S_n/S_{pi}]$ decomposes into simples, since they all get lumped together.