 Solution

Assume $n\ge 2$. The place-permutation action on ${\mathbb{C}}^n$ preserves the coefficient-sum kernelIt 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 $T_j$, $2\le j\le n$, with $j$ below the first cell and the remaining entries increasing along the first row. Their content vectors of standard Young tableaux areAn explicit orthonormal basis realizing these tableau lines isThe 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 $(ar)$ gives $X_r{w}_j=c_{T_j}(r)w_j$.

For $s_i=(ii+1)$, $s_1{w}_2=-w_2$. For $2\le i\le n-1$, its only nontrivial two-dimensional block isEvery 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 $T_i$.