= Solution
Write the <finite abelian group> additively. Its <group shift operator> is
$$
\boxed{U(h)|g\rangle=|g+h\rangle.}
$$
These <unitary operators> form the <regular representation> and commute. The representation-theoretic facts we use are that every <irreducible representation of a finite abelian group> is one-dimensional, there are $|G|$ such <characters of a representation>, and their <character orthogonality> gives an <orthonormal basis> of functions on $G$. Each <linear character> here is a <group homomorphism> $\chi:G\to U(1)$, with $|\chi(g)|=1$.
For $\chi\in\widehat G$, the <character group of a finite abelian group>, put
$$
|v_\chi\rangle=\frac1{\sqrt{|G|}}\sum_{g\in G}\overline{\chi(g)}|g\rangle.
$$
Changing variables to $u=g+h$ gives
$$
U(h)|v_\chi\rangle
=\frac1{\sqrt{|G|}}\sum_u\overline{\chi(u-h)}|u\rangle
=\boxed{\chi(h)|v_\chi\rangle}.
$$
The $|G|$ vectors $|v_\chi\rangle$ are therefore a common <eigenbasis>. Using $\chi(g)$ instead of its <complex conjugate> in their definition reverses every <eigenphase>; this is merely the <quantum Fourier transform> sign convention.
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