 Cyclic eigenvector obstruction to invariant vectors

In a unitary representation of a finite group, if a cyclic vector $v$ satisfies $\rho (g)v=av$ for some $a\ne 1$, then there are no invariant vectors. Indeed the averaging orthogonal projection $P=|G{|}^{-1}{\sum }_h\rho (h)$ satisfies $P\rho (g)=P$, so $Pv=aPv=0$. All translates of $v$ also project to zero, and they span the representation.