= Solution
Put $m=n-k$ and take the subgroup $S_m\times S_k$ permuting the first $m$ and last $k$ letters. Define the <skew representation of a symmetric group> as the multiplicity space
$$
\boxed{V^{\lambda/\mu}=\operatorname{Hom}_{S_m}(V^\mu,\operatorname{Res}^{S_n}_{S_m}V^\lambda).}
$$
The last-letter copy of $S_k$ commutes with $S_m$, so it acts on a map $f$ by $g\cdot f=\rho_\lambda(g)\circ f$. This gives a genuine $\mathbb C S_k$-module, without claiming that the whole restricted module is itself the skew representation. Equivalently,
$$
\operatorname{Res}^{S_n}_{S_m\times S_k}V^\lambda\cong\bigoplus_{\nu\vdash m}V^\nu\otimes V^{\lambda/\nu}.
$$
Fix a prefix tableau of shape $\mu$ and complete it to shape $\lambda$. Iterated <restriction branching rule for a symmetric group> identifies the multiplicity-space <orthonormal basis> with <standard skew Young tableaux>, using labels $1,\ldots,k$ for the last $k$ cells. The <Young orthogonal form> restricts to this basis. When $R=s_iT$ is standard and $d=c_T(i+1)-c_T(i)$,
$$
s_iw_T=d^{-1}w_T+\sqrt{1-d^{-2}}\,w_R.
$$
An admissible interchange has $|d|\geq2$, so its off-diagonal coefficient is nonzero. Consequently $w_R$ belongs to the <group algebra> span of $w_T$. The preceding reduced-path argument reaches every standard skew tableau, so this span is the whole module. \b[Every $w_T$ is a <cyclic vector for a group representation>.] The inherited invariant inner product makes this a <unitary representation>.
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