The measure-equivalence conjecture for words in free groups

Let $F$) be a finitely generated free group, and let $u,w\in F$ be words. Two words are measure equivalent in all finite groups if they have equally sized fibres under their word maps in every finite group.

Measure-equivalence conjecture. If $u$ and $w$ are measure equivalent in all finite groups, then there exists $\phi\in\operatorname{Aut}(F)$ such that

$$u=\phi(w).$$

The conjecture is known when one word is primitive or a surface word; the paper also proves the surface-word case using only finite solvable quotients. The general case remains open.