Suppose $z$ is drawn from a continuous distribution on $[0,1]$ with c.d.f. $F(z)$.
Suppose that given $z$, $x$ is drawn from a Beta distribution with $\alpha=z\nu$ and $\beta=(1-z)\nu$, where $\nu>0$.
Then the unconditional distribution of $x$ has p.d.f.:
$$\int_0^1 { \frac{x^{z\nu-1}(1-x)^{(1-z)\nu-1}}{B\left(z\nu,(1-z)\nu\right) } \, dF(z)}$$
where $B$ is the Beta function.
My question is whether there is some $F$ such that the resulting distribution is uniform.
In other words: find $F$ to solve the integral equation:
$$\int_0^1 { \frac{x^{z\nu-1}(1-x)^{(1-z)\nu-1}}{B\left(z\nu,(1-z)\nu\right) } \, dF(z)}=1$$
(Note: The solution for $F$ will of course depend on $\nu$.)
Some limited progress towards a solution:
Suppose $\nu=2$. Then $F$ equal to a step function with the step (from $0$ to $1$) located at $z=\frac{1}{2}$ is a solution.
As $\nu$ increases, the variance of $x\mid z$ decreases, thus $z$ needs to become more dispersed. This suggests that no solution exists for $\nu<2$.