 Solution

The symmetric group is transitive on two-element subsets, giving $⟨1,{\chi }_Y⟩=1$. Its diagonal action on $X\times Y$ has two orbits, according to whether the point lies in the pair. Thus $⟨{\chi }_X,{\chi }_Y⟩=2$, and, since ${\chi }_X=1+{\chi }_S$, the standard zero-sum representation $S$ occurs in $\mathbb{C}[Y]$ once. Its irreducibility follows from the double transitivity of the natural symmetric-group action and the corrected statement of part (ii).

For $n\ge 4$, ordered pairs of two-element subsets have three possible intersection sizes, zero, one and two, each forming a single orbit. Part (i) therefore gives $\Vert {\chi }_Y{\Vert }^2=3$. After removing the one copy each of $1$ and $S$, the remaining character has squared norm $3-1-1=1$, so is itself irreducible. HenceFor $n=3$, taking the complement of a pair is an equivariant bijection $Y\to X$. Thus $\mathbb{C}[Y]\cong 1\oplus S$, with no third constituent. Correspondingly the disjoint-pair orbit is absent and the character norm is two.