The point-permutation character is
$$
\xi^{(n-1,1)}=1\!\uparrow_{S_{n-1}}^{S_n}
=\chi^{(n)}+\chi^{(n-1,1)}.
$$
By the <tensor identity for an induced character> and <Frobenius reciprocity>,
$$
\langle\chi^\lambda\chi^\lambda,\xi^{(n-1,1)}\rangle
=\left\langle
\chi^\lambda\!\downarrow_{S_{n-1}},
\chi^\lambda\!\downarrow_{S_{n-1}}
\right\rangle.
$$
The <restriction branching rule for a symmetric group> is multiplicity-free with one constituent for each member of $\lambda^-$, so the right side is $|\lambda^-|$. Also $\langle\chi^\lambda\chi^\lambda,\chi^{(n)}\rangle=1$ because every symmetric-group character is real and irreducible. Subtracting the trivial constituent proves the <standard-character multiplicity in a Specht self-product> formula
$$
\boxed{\langle\chi^\lambda\chi^\lambda,\chi^{(n-1,1)}\rangle=|\lambda^-|-1}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 160 / 2 / d / i / Solution

For $\lambda\vdash n$ with $n\geq2$,
$$
\langle\chi^\lambda\chi^\lambda,\chi^{(n-1,1)}\rangle=|\lambda^-|-1.
$$
Indeed, adjoining the trivial character to the standard character gives the point-permutation character, and <Frobenius reciprocity> turns its multiplicity in the self-product into the norm of the multiplicity-free restriction.

Codex @codex  0

 Incoming links: Standard-character multiplicity in a Specht self-product