Question:
Let $u,v \in C^1 ([0,1], \mathbb{R}^+)$ such that $u$ is increasing and $v$ is deacreasing and $u(0)=v(1)=0$. Suppose furthermore that the real function $0 \le x(t)\le 1$ satisfies $x'=(1-x)u(x) - xv(x)$. From here can we deduce that the system admits a unique stationary point other than the trivial $x=1$ and $x=0$?
Under the stated assumptions on $u$, $v$ we cannot deduce the existence of another stationary point. Also if such a point exists it does not have to be unique.
For the first claim (the existence of stationary points in $(0,1)$ is not a necessary condition) consider for example $u(x)=x^2$ which is strictly increasing with $u(0)=0$ and $v(x)=1-x$ which is strictly decreasing with $v(1)=0$. Then, $$x'=(1-x)u(x)-xv(x)=(1-x)x^2-x(1-x)=-x(1-x)^2$$ for which the only stationary points are $x=0$ and $x=1$.
For the second claim the obvious selection $u(x)=x$, $v(x)=1-x$ yields an infinite number of stationary points (all points in $[0,1]$).