How to map $\alpha x$ and $(1 - \alpha)x$ for $\alpha \in (0,1)$ to the same point Can you please help me with a problem from real function analysis? 
Let $p$ be a continuous, increasing function such that $p(s) \in (0,1)$ for $s > 0$. Let $f$ and $g$ be two functions defined on $\mathbb{R}_+$ such that $f(s) = s p(s)$ and $g(s) = s - f(s) = s(1-p(s))$. I am trying to find two nondecreasing functions $F$ and $G$ such that $F(f(s)) = G(g(s))$ for $s > 0$.
We can easily see that $0 < f(s) < s$, $0 < g(s) < s$, and for example that $|f(s) - \frac{s}{2}| = |g(s) - \frac{s}{2}|$.
One solution is $F = f^{-1}$ and $G = g^{-1}$ assuming that $p$ is such that $f$ and $g$ are both nondecreasing, their inverses exist (what else?). Such $F$ and $G$ work fine for, for example, $p(s)=s/(1+s)$, but not for $p(s) = s^2/(1+s^2)$, since $g$ is not increasing for all $s > 0$ in the latter case.
However, I cannot find any $F$ and $G$ that would work for a general increasing function $p$. (One of them, either $F$ or $G$, can be Identity, if it helps). Any ideas?
Many thanks in advance!
EDIT: As pointed out by Adam Latosiński, the answer is negative for general $p$. Is it possible to find $F$ and $G$, other than inverses of $f$ and $g$, in the case when $p$ is increasing and such that $f$ and $g$ are increasing as well?
 A: In general it's not possible. For example, if $f$ is nondecreasing, but $g$ is not (at least not always) then for nondecreasing functions $F$ and $G$ we'll have that $F\circ f$ is be nondecreasing, but $G\circ g$ is not nondecreasing, Therefore these functions cannot be equal.
If $f$ and $g$ are nondecreasing and $f^{-1}$ and $g^{-1}$ exist, then from the condition
$$ F(f(s)) = G(g(s)) =: h(s)$$
you have 
$$ F(t) = h(f^{-1}(t)) \\ G(t) = h(g^{-1}(t)) $$
Using various non-decreasing functions $h$ will give you various functions $F$ and $G$.
If $f$ and $g$ are nondecreasing but at least one of $f^{-1}$ and $g^{-1}$ does not exist (which means that there are intervals on which they are constant) we can still use similar construction. From the condition $$ F(f(s)) = G(g(s)) =: h(s)$$ it follows that $$ t = f(s) \qquad \Rightarrow \qquad F(t) = h(s)$$ which means that for $s_1$ and $s_2$ such that $  f(s_1)=f(s_2) $ we need $ h(s_1) = h(s_2)$. Similarily, if  $g(s_1)=g(s_2) $ we need $ h(s_1) = h(s_2)$. That limits our choice of function $h$ to such functions that are constant on every interval on which $f$ or $g$ are constant. When we pick any such function, then the condition $ F(f(s)) = G(g(s)) = h(s)$ defines functions $F$ and $G$ uniquely. For example, if $f^{-1}$ exists but $g^{-1}$ doesn't, you can take $F=g\circ f^{-1}$, $G={\rm id}$.
