The conditions on $F$ are
$F(0)=0$
$F(\overline{\alpha}) = 1$
$F'(x) \gt 0$ for all $x \in (0, \overline{\alpha})$
$\int_0^{\overline{\alpha}} x F'(x) dx = 1$.
$F'$ is differentiable.
These are all nonlocal in the sense that sufficiently small (smooth) perturbations of $F$ within small neighborhoods of any finite discrete set of points in $(0, \overline{\alpha})$ can be found which preserve them all. By applying such a local perturbation we can modify $F'$ within a narrow interval to be as large as we want without changing any of the conditions.
I claim there exists at least one $x_0 \in (0, \overline{\alpha})$ for which $(N-1)(1-\delta+\delta F(x_0)) - \delta x_0 F'(x_0) \gt 0$. For if not, a simple comparison shows that $F(0)$ must be less than the value at $0$ of the (unique) solution to the first-order ordinary differential equation for $F$,
$$L[F](x) = (N-1)(1-\delta+\delta F(x)) - \delta x F'(x) = 0, \quad F(\overline{\alpha}) = 1.$$
This equation can explicitly be solved and its value at $0$ found to equal $1 - 1/\delta \lt 0$, whence $F(0) \lt 0$, violating the first condition.
Let's perturb $F$ to $\widehat{F}$. The following argument needs only routine constructions to be made rigorous: mix in tiny amounts of a distribution supported in an arbitrarily small neighborhood of $x_0$ and another distribution supported in a neighborhood of an $x_1$ on the other side of $1$ from $x_0$ in such a way that the expectation of $F$ remains unchanged. By making the first of these mixed-in distributions have sufficiently small support, we can cause it to increase $F'$ at $x_0$ by any desired amount while changing $F(x_0)$ arbitrarily little. In this fashion we can cause $L[\widehat{F}](x_0)$ to range from $L[F](x_0)$ down to $-\infty$ in a continuous manner. Do this so that $L[\widehat{F}](x_0) = 0$.
Therefore, there exist no restrictions whatsoever on any of the moments of $F$. That explains why you had such difficulty obtaining any!