 Solution

The permutation representation on the complex vector space with basis $X$ has permutation character ${\pi }_X(g)=|{\mathrm{Fix}}_X(g)|$. The multiplicity of the trivial character is $⟨{\pi }_X,1_G⟩=|G{|}^{-1}{\sum }_g|\mathrm{Fix}(g)|$. Count pairs $(g,x)$ with $gx=x$: each orbit contributes $|G|$, by the orbit-stabilizer theorem. Hence this multiplicity equals the number of orbits.

For a transitive action the average number of fixed points is one. The identity fixes $|X|>1$ points, so not every other element can fix at least one; otherwise the average would exceed one. Thus there is a derangement $g$ with ${\pi }_X(g)=0$.

If ${\pi }_X=1_G+m\chi$, evaluating at that derangement gives $\chi (g)=-1/m$. Character values are algebraic integers, since a finite-order representation matrix has roots of unity as eigenvalues, and the sum of algebraic integers is integral. A rational algebraic integer is an integer. Therefore $-1/m$ is an integer, forcing $m=1$.

The diagonal action on $X^2$ is $g(x,y)=(gx,gy)$. Its permutation character is ${\pi }_X(g{)}^2$, since both coordinates must be fixed. The orbit-count result and the real-valued character ${\pi }_X$ giveIf any $m_j\ge 2$ and another nontrivial constituent occurs, this is at least $1+4+1=6$. If just one nontrivial constituent occurs, the previous result forces its multiplicity to one. Thus $r\le 5$ implies every multiplicity is one, and the formula then gives $k=r$.