We say an invariant $P$ is a $\textit{complete invariant}$ for an equivalence relation $\sim$ if whenever $T$ and $S$ both have the property $P$, then $T \sim S$. By a Theorem of Ornstein (see section $4.9$ of Walters' "An Introduction to Ergodic Theory") we know that if $T_{1}$ and $T_{2}$ are two Bernoulli shifts whose state spaces are Lebesgue spaces, and $h(T_{1}) = h(T_{2})$, then $T_{1}$ is conjugate to $T_{2}$ (here $h$ denotes measure-theoretic entropy and by conjugate I mean conjugacy of measure-preserving transformations). Thus entropy is a complete invariant for the relation of conjugacy on the class of Bernoulli shifts with state spaces Lebesgue spaces.
However, the same is not true for rotations of compact abelian groups. Indeed, consider two rotations of the circle $T_{1}(z) = az$ and $T_{2}(z) = bz$ where $\{a^{n} : n \in \mathbb{Z} \}$ and $\{b^{n} : n \in \mathbb{Z} \}$ are distinct dense subsets of the circle. Then $T_{1}$ and $T_{2}$ are ergodic measure-preserving transformations with discrete spectrum, and $h(T_{1}) = 0 = h(T_{2})$. However since their eigenvalues (by which I mean the eigenvalues of the associated linear operators $U_{T_{i}} : L^2 (m_{i}) \rightarrow L^{2}(m_{i}), f \mapsto f \circ T_{i}$) are distinct by construction, $T_{1}$ is not conjugate to $T_{2}$. Hence entropy is not a complete conjugacy invariant for such transformations.
My question is then: is there a known complete conjugacy invariant for rotations of compact abelian groups?
Thank you in advance!