= Solution
Work with the selected transitive orbit $X\cong G/H$. By <Maschke's theorem>, decompose its permutation character as
$$
1_H\mathbin{\uparrow}^G
=1_G+\sum_jm_j\chi_j,
$$
where the $\chi_j$ are distinct nontrivial <irreducible characters>. <Character orthogonality> gives
$$
\left\langle1_H\mathbin{\uparrow}^G,
1_H\mathbin{\uparrow}^G\right\rangle_G
=1+\sum_jm_j^2.
$$
Consequently the augmentation summand $V$ is <irreducible> exactly when this inner product equals $2$.
The <double cosets> $H\backslash G/H$ are the $H$-orbits on $G/H$. Because $G$ acts transitively on $X$, these in turn correspond to the $G$-orbits on $X\times X$: move the first coordinate to the base point $H$, after which its stabilizer is $H$. One orbit is the diagonal. There are exactly two orbits precisely when $H$ is transitive on $X\setminus\{H\}$, equivalently when the action is <two-transitive>. Part (ii) therefore proves
$$
\boxed{V\text{ is irreducible}
\quad\Longleftrightarrow\quad
G\text{ acts two-transitively on }X.}
$$
This is the <irreducible augmentation criterion for a transitive group action>.
Finally, transitivity implies that the fixed-vector space $\mathbb C[X]^G$ consists only of constant vectors and has dimension one. If the nonzero representation $V$ were a <trivial representation>, it would contribute another fixed vector, a contradiction. Thus $V$ is not the trivial representation.
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