= Solution
Assume $n\geq2$. The place-permutation action on $\mathbb C^n$ preserves the coefficient-sum kernel
$$
W=\{z:\textstyle\sum_jz_j=0\}.
$$
It is the <augmentation subrepresentation of a permutation representation>. The point action of $S_n$ is <two-transitive>, so the <irreducible augmentation criterion for a transitive group action> makes $W$ irreducible. The <two-row Young permutation module decomposition> of $M^{(n-1,1)}\cong\mathbb C^n$ identifies its nontrivial summand as $V^{(n-1,1)}$. Hence $W\cong V^{(n-1,1)}$.
All <standard Young tableaux> of shape $(n-1,1)$ are $T_j$, $2\leq j\leq n$, with $j$ below the first cell and the remaining entries increasing along the first row. Their <content vectors of standard Young tableaux> are
$$
c_{T_j}(r)=\begin{cases}r-1&r<j,\\-1&r=j,\\r-2&r>j.\end{cases}
$$
An explicit <orthonormal basis> realizing these tableau lines is
$$
w_j=\frac{e_1+\cdots+e_{j-1}-(j-1)e_j}{\sqrt{j(j-1)}}.
$$
The sums of their coordinates vanish. Their norms are one, and the inner product of $w_j$ with $w_l$, $j<l$, is zero because the coefficients of $w_j$ sum to zero. Directly summing the action of $(a\ r)$ gives $X_rw_j=c_{T_j}(r)w_j$.
For $s_i=(i\ i+1)$, $s_1w_2=-w_2$. For $2\leq i\leq n-1$, its only nontrivial two-dimensional block is
$$
\boxed{\begin{aligned}
s_iw_i&=\frac1i w_i+\sqrt{1-\frac1{i^2}}w_{i+1},\\
s_iw_{i+1}&=\sqrt{1-\frac1{i^2}}w_i-\frac1i w_{i+1}.
\end{aligned}}
$$
Every other $w_j$ is fixed, including all $w_j$ with $j>2$ when $i=1$. These formulas follow by swapping coordinates $i,i+1$ in the displayed vectors, and are the <Young orthogonal form> with axial distance $i$ for $T_i$.
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