 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 andThis is the abacus divisible-hook correspondence.