= Solution
Let $P$ be the averaging <linear map>
$$
P=\frac1{|G|}\sum_{h\in G}\rho(h).
$$
For a <unitary representation> of a <finite group>, $P$ is the <orthogonal projection> onto the <invariant subspace> $V^G$, and $P\rho(h)=\rho(h)P=P$. Applying this to $\rho(g)v=\lambda v$ gives $Pv=\lambda Pv$. Since $\lambda\ne1$, $Pv=0$. The vector $v$ is a <cyclic vector for a group representation>, so its translates span $V$, and every translate also has projection zero. Thus $P=0$ and
$$
\boxed{V^G=\{0\}.}
$$
This is the <cyclic eigenvector obstruction to invariant vectors>. Unitarity also gives a direct proof by pairing $v$ with an invariant vector and then pairing every translate; the averaging proof makes the role of cyclicity especially transparent.
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