 Solution

Assume $n\ge 2$. The standard Young tableaux of shape $(n-1,1)$ are $T_k$, for $2\le k\le n$: the sole second-row entry is $k$, and the first row isThis lists all of them, since $1$ must occupy the first box and the rest of the first row is forced to increase. Their contents areWrite $w_k=w_{T_k}$. The only nontrivial adjacent swaps exchange the lower entry with $k-1$ or $k+1$. Substituting the content differences into Young orthogonal form yields all generator actions:In the first line at $k=2$, the square-root coefficient is zero and the action is $-w_2$, so no vector $w_1$ is needed. The second line is used only when $k<n$, since the generator $s_n$ does not exist.

The corresponding Specht module is the sum-zero standard representation of the symmetric group:A concrete orthonormal realization isPermuting coordinates gives exactly the actions above, linking this orthogonal tableau basis with the Gelfand–Tsetlin basis found in Question 1. Here the printed $s_j$ are the adjacent Coxeter generators, rather than products of all generators.