Given a funtion $f \in L^p([0,1])$ (take $p=\infty$ if you'd like), and also a measure preserving map $s:[0,1] \to [0,1]$ (meaning $s$ pushes Lebesgue measure forward to itself) I would like to know if there exists some $f^{\ast} \in L^p([0,1])$ such that $f^{\ast} \circ s = f$. If $s$ is invertible this is of course obvious but measure preserving maps need not be invertible (although must be onto).

Recall that given $f$ there exists a monotone rearrangement of $f$, and a measure preserving map $t:[0,1] \to [0,1]$ which yields this rearrangement.

However my question is in some sense the reverse question (with no monotonicity added).

Somehow it seems intuitive that such a map should exist but I'm not able to prove it directly. It seems like something which may be well known however.