I have (expect there exists) a function $f: (0,1) \to \mathbb R$ for which the following relationship holds: $$ f\left(\frac{2-p+p^2}{4-2p}\right) - f(p/2) =f\left(\frac{(1-p)(2+p)}{2-p}\right) - f(1-p), $$ for all $p\in(0,1)$. (Possibly there is a nice choice of $p$ that makes this look nicer.)
I know the function is smooth, monotonously increasing, and goes to infinity as its argument goes to 1 (respectively to $-\infty$ when the argument goes to 0.) That is, it's the inverse of a sigmoid function.
I think the above characterization should be enough to determine the function, at least up to some scaling and translation (if $f(x) = a g(x) + b$ clearly $a$ and $b$ cancel.)
Besides that I have no idea about how I might go about finding $f$. I don't even know how I could try to compute it numerically. Any help greatly appreciated.
Edit: Maybe it's nicer to state like this: $$ \begin{align} &f\left(p_1\frac{1-p_2p_3}{(1-p_2)(1-p_3)}\right)-f(p_1) \\ =&f\left(p_2\frac{1-p_1p_3}{(1-p_1)(1-p_3)}\right)-f(p_2) \\ =&f\left(p_3\frac{1-p_1p_2}{(1-p_1)(1-p_2)}\right)-f(p_3) \end{align} $$ for any choice of $p_1,p_2,p_3\ge 0, p_1+p_2+p_3=1$.
The formulation above comes from considering $p_1=p_2=p/2$ and $p_3=1-p$.
At least your “nicer” version has no solution. At $p_1 = 0.6, p_2 = 0.03, p_3 = 0.37$, we have
$$f(0.970938) - f(0.6) = f(0.092619) - f(0.03).$$
At $p_1 = 0.03, p_2 = 0.092619, p_3 = 0.877381$, we have
$$f(0.247722) - f(0.03) = f(0.758204) - f(0.092619).$$
At $p_1 = 0.247722, p_2 = 0.662382, p_3 = 0.089896$, we have
$$f(0.758204) - f(0.247722) = f(0.945929) - f(0.662382).$$
Adding these three equations and canceling yields
$$f(0.970938) - f(0.6) = f(0.945929) - f(0.662382).$$
But for $f$ to be strictly increasing, the left side must be greater than the right side.