= Solution
Using the <abacus of a partition> in James's convention, read runner $r$ as a one-runner abacus: replace each occupied position $r+eq$ by $q$. After the usual harmless shift of its beta set, the resulting partition is $\lambda^{(r)}$. Doing this for $r=0,\ldots,e-1$ produces
$$
Q_e(\lambda)=(\lambda^{(0)},\ldots,\lambda^{(e-1)}).
$$
Conjugating a partition complements beads and gaps and reflects the abacus. Reflection sends runner $r$ to runner $e-1-r$, reverses bead-gap order, and conjugates the runner partition. Hence
$$
Q_e(\lambda')
=\bigl((\lambda^{(e-1)})',\ldots,
(\lambda^{(1)})',(\lambda^{(0)})'\bigr).
$$
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