= Solution
For a <standard Young tableau> $T$, let $c_T(r)=j-i$ when its entry $r$ lies in cell $(i,j)$. Its <content vector of a standard Young tableau> is $C(T)=(c_T(1),\ldots,c_T(n))$. We use the following sign convention for <axial distance in a Young tableau>:
$$
\boxed{d_T(i,j)=c_T(i)-c_T(j)}.
$$
This is the axial distance from $j$ to $i$. In particular, write $d=c_T(r+1)-c_T(r)$ for the adjacent <transposition> $s_r=(r\ r+1)$.
Here is <Young orthogonal form>, with one <orthonormal basis> vector $w_T$ for each <standard Young tableau> of a fixed shape:
$$
\boxed{s_rw_T=\frac1d\,w_T+\sqrt{1-\frac1{d^2}}\,w_{s_rT}}.
$$
If $s_rT$ is not standard, omit that term. In this case $r,r+1$ occupy adjacent cells in one row or column, so $d=1$ or $d=-1$, respectively, and the scalar action is $+1$ or $-1$. On a standard pair $T,R=s_rT$, $|d|\geq2$ and the matrix is
$$
\begin{pmatrix}d^{-1}&\sqrt{1-d^{-2}}\\
\sqrt{1-d^{-2}}&-d^{-1}\end{pmatrix}.
$$
This is a real orthogonal involution.
To prove the formula, explicitly use the permitted <Young seminormal form> theorem: the <Specht module> has a tableau basis $v_T$ whose vectors are joint eigenvectors of the <Young–Jucys–Murphy elements>, and whose actions, on an admissible pair oriented so that tableau length increases from $T$ to $R$, are
$$
s_rv_T=d^{-1}v_T+v_R,\qquad
s_rv_R=(1-d^{-2})v_T-d^{-1}v_R.
$$
Nonadmissible swaps have the scalar actions just described. The theorem supplies these as one globally consistent representation, not as independently chosen two-dimensional blocks.
Average a positive definite <Hermitian inner product> over $S_n$. Distinct tableau lines are orthogonal because their commuting self-adjoint <Young–Jucys–Murphy elements> have distinct joint weights. Also $s_r$ is self-adjoint, so the two seminormal equations imply
$$
\|v_R\|^2=(1-d^{-2})\|v_T\|^2.
$$
Upon setting $w_T=v_T/\|v_T\|$, the coefficient from $w_T$ to $w_R$ becomes $\|v_R\|/\|v_T\|=\sqrt{1-d^{-2}}$; the reverse coefficient is the same. All phases are consistent because the seminormal basis was global, and the group relations remain true under this change of basis. This proves <Young orthogonal form>.
For $\lambda=(n)$, there is exactly one <standard Young tableau>, the row $1,2,\ldots,n$. Its <content vector of a standard Young tableau> is
$$
\boxed{C(T)=(0,1,\ldots,n-1)}.
$$
Every adjacent distance is $1$, so \b[$s_rw_T=w_T$ for every $r$]. Thus $\boxed{S^{(n)}\cong\mathbb C_{\rm triv}}$ is the <trivial representation>.
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