= Solution
The <Gelfand–Tsetlin basis> spans $V^\lambda$, and the scalar computed on its vectors depends only on the shape. Thus
$$
\boxed{X_2\cdots X_nw=\begin{cases}(-1)^k k!(n-k-1)!\,w&\lambda=(n-k,1^k),\\0&\lambda\text{ not a hook},\end{cases}}
$$
for every $w\in V^\lambda$. The product is the sum of the $(n-1)!$ permutations in the <conjugacy class> of an $n$-cycle. Taking <traces> therefore gives $(n-1)!\chi^\lambda((n))$ equal to the displayed scalar times $\dim V^\lambda$.
For a <hook partition>, a <standard Young tableau> is uniquely determined by the choice of its $k$ entries below the top cell, selected from $\{2,\ldots,n\}$. The column and the remaining row are then forced to increase. Hence $\dim V^{(n-k,1^k)}=\binom{n-1}{k}$, and cancellation of the factorials yields
$$
\boxed{\chi^\lambda((n))=\begin{cases}(-1)^k&\lambda=(n-k,1^k),\\0&\lambda\text{ not a hook}.\end{cases}}
$$
This uses the <central character value of a conjugacy-class sum> and tableau counting, without a character rule for removing <rim hooks>.
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