= Solution
Use a <Gelfand–Tsetlin basis> in <Young seminormal form>.
For a <standard Young tableau> $T$, put $d=c_T(i+1)-c_T(i)$, using the <Content of a Young-diagram cell>. Choose the row-reading tableau $T_0$, and let $\ell(T)$ be the <Coxeter length> of the unique permutation sending $T_0$ to $T$. A <Gelfand–Tsetlin basis> can be chosen so that, when $R=s_iT$ is standard and $\ell(R)=\ell(T)+1$,
$$
\boxed{s_iv_T=d^{-1}v_T+v_R,\qquad
s_iv_R=(1-d^{-2})v_T-d^{-1}v_R.}
$$
If $R$ is not standard, the action is $+v_T$ for two consecutive entries in one row, and $-v_T$ for two in one column. This is one usual normalization of the <Young seminormal form>.
Here is a construction and proof of the normalization. Fix $v_{T_0}\ne0$, let $P_T$ be the projection onto the tableau line, and define
$$
v_T=P_T\pi_Tv_{T_0}.
$$
The permutation $\pi_T$ has a reduced expression consisting entirely of admissible swaps, by the <reduced adjacent-swap path between linear extensions>. At each swap the off-diagonal coefficient is nonzero. In its expansion, the only term that can reach a tableau at distance $\ell(T)$ uses all $\ell(T)$ swaps; omitting a swap gives a shorter path. Thus $v_T\ne0$. This also makes its definition independent of a chosen reduced expression, because $\pi_T$ itself is fixed.
The relation $s_iX_is_i+s_i=X_{i+1}$ forces the coefficient of $v_T$ in $s_iv_T$ to be $1/d$, and forces every other component to lie on the swapped line. If length increases, project the identity $\pi_R=s_i\pi_T$ onto the line for $R$. The shorter terms of $\pi_Tv_{T_0}$ cannot reach $R$ in one step, so the coefficient of $v_R$ is exactly one. Applying $s_i^2=1$ then gives the reverse coefficient $1-d^{-2}$ and diagonal coefficient $-1/d$. In the nonstandard cases the same relation gives $d=\pm1$ and the asserted scalar action. This proves the theorem rather than merely specifying pairwise scalings.
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