Not an answer by any means, but too long for a comment.

Assume first $X$ is closable, thus $X_t = \mathbb E[X_1 | \mathcal F_t]$, where $\mathcal F_t$ is the natural filtration of $X$. Assume also we are lucky enough that there exists a Monge map $T$ transporting $\mu$ onto $\nu$, that is $T_* \mu = \nu$. 

Then if this result is possible at all, it *has* to hold (read disclaimer!) in this nice case that the martingales $X_t$ and $Y_t := \mathbb E[T(X_1)| \mathcal F_t]$ are what you’re looking for.

**Disclaimer:** Obviously I am not 100% sure! But this is the nicest case I could think of, which could serve as a good test case. In the case where there isn’t a Monge map, one could replace $T$ with some optimal coupling and see if it works.

**Edit:** Actually does my claim follow from Doob’s inequalities? Don’t have time to work it out in detail for now…