= Solution
The <permutation representation> on the complex <vector> space with basis $X$ has <permutation character> $\pi_X(g)=|\operatorname{Fix}_X(g)|$. The multiplicity of the trivial character is $\langle\pi_X,1_G\rangle=|G|^{-1}\sum_g|\operatorname{Fix}(g)|$. 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 \b[$m=1$].
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 $\pi_X$ give
$$
\boxed{r=\frac1{|G|}\sum_g\pi_X(g)^2=\langle\pi_X,\pi_X\rangle
=1+\sum_{j=2}^km_j^2.}
$$
If any $m_j\geq2$ 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 $r\leq5$ implies every multiplicity is one, and the formula then gives \b[$k=r$].
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