 Solution

As printed, the equivalence needs a transitivity hypothesis. A group acting trivially on two points has $\mathbb{C}\Omega ={\mathbb{C}}_G\oplus {\mathbb{C}}_G$, so $V$ is one-dimensional and irreducible, although the action is not even transitive.

Here is the intended result for a transitive action. The set is then finite, and, for a point stabilizer $H$, its permutation character is $\chi ={\mathrm{Ind}}_H^{G}1$. Frobenius reciprocity givesThe last equality follows because an $H$-fixed vector has constant coefficients on each orbit. There is exactly one trivial constituent, since the action is transitive. Therefore, writing $\chi =1+\psi$ for the character of $V$,By character orthogonality, $V$ is irreducible exactly when this equals $1$. One stabilizer orbit is $\{\omega \}$, so this happens exactly when $H$ is transitive on $\Omega \setminus \{\omega \}$, equivalently when $G$ is two-transitive. Moreover $V$ is not trivial, since a transitive action has only one independent fixed vector in the whole permutation module.

The complete correction without assuming transitivity is also precise: $V$ is irreducible exactly for a two-transitive action or the exceptional action on two fixed points. Indeed for a finite set with $r$ orbits, the fixed subspace of $V$ has dimension $r-1$. If $r>1$ and $V$ is irreducible, $V$ must itself be a one-dimensional trivial module, forcing $|\Omega |=2$ and $r=2$. If $\Omega$ is infinite, $V$ is infinite-dimensional and cannot be irreducible: any nonzero vector generates a submodule of dimension at most $|G|$.