It is known that minimality does not imply unique ergodicity (Furstenberg example). I ask whether the implication holds in following particular situation:

Suppose $X$ is a compact space, $f:X \to X$ is a minimal and uniquely ergodic homeomorphism. Suppose $p: Y \to X$ is a

2-to-1 covering map, and that $g:Y \to Y$ is a homeomorphism covering $f$, i.e., $p \circ g = f \circ p$. Suppose $g$ is minimal.Does it follow that $g$ is uniquely ergodic?

Trivial example: If $f$ is an irrational rotation of the circle then it is easy to see that so is $g$, and so in this case the question has a positive answer.

A less general but already interesting situation is when the covering is trivial, i.e., $Y = X \times \{0,1\}$. In this case, I think that my question is equivalent to the following **coboundary rigidity** question:

Let $f$ be as above. Let $\phi:X \to \mathbb{Z}_2$ be a continuous map

on the group with two elements. Suppose it is a measurable-coboundary, i.e., there exists a measurable map $\psi:X \to \mathbb{Z}_2$ such that $\phi = \psi \circ f - \psi$ almost everywhere (w.r.t. the unique $f$-invariant probability measure).Does it follow that $\phi$ is a continuous-coboundary?(I.e., can $\psi$ be chosen continuous?)

Rem.: To relate the two questions, notice that $g(x,t)=(f(x),t+\phi(x))$ is uniquely ergodic (resp., minimal) iff $\phi$ is not a measurable-coboundary (resp., not a continuous coboundary).