Showing an ergodic toral automorphism is not measurably isomorphic to an ergodic circle rotation

Question

The question as listed in the title is the question statement, only I do not want to use that one is mixing and the other is not.

Is it true that measurably isomorphic spaces are either both mixing or both non-mixing? If so, can you be so kind to tell why?

I was wondering if a direct proof to the question in the title exists that does not use mixing.

Answer 1

To show that they are not isomorphic it is fine to give a metric invariant that is different in the two.

For example: periodic points are very different in both cases, but the infinite set of periodic for a toral automorphism still has zero measure; I personally like mixing, but you may take entropy: one is positive and the other is zero.

PS: I wonder whether you wanted to write "hyperbolic toral automorphism" instead of "ergodic toral automorphism" since otherwise some technicalities need to be dealt with.

Added: The most elementary approach seems to be showing that if $T$ is the toral automorphism and $S$ is the circle rotation, then $T\times T$ is ergodic but $S\times S$ is not ergodic, with the corresponding Lebesgue measures.