= Solution
Use a <beta set> $X$ on an $e$-runner <partition abacus>. A hook of length divisible by $e$ corresponds to a bead $x$ and a gap $x-eh$ on the same runner. Divide both runner positions by $e$. They become a bead and a gap at distance $h$ in the runner partition $\lambda^{(s)}$, and hence determine a hook there. This gives the required <bijection>, with $|H|=e|f(H)|$.
Removing $H$ replaces the bead $x$ by the gap $x-eh$. On its runner this is exactly the bead move that removes $f(H)$, while every other runner is unchanged. Thus hook removal commutes with the construction and
$$
Q_e(\lambda\setminus H)=Q_e(\lambda)\setminus f(H).
$$
This is the <abacus divisible-hook correspondence>.
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