It started where the expression $\frac{1-x}{x^2}$ came up, and in thinking to simplify the system of equations, I thought there might be a substitution I could make: "what if there is a constant $a$ such that, for $h(x) = \frac{1-x}{x^2},$ $h(x-a) = \pm \frac{1}{x-a}$?"
Well, this didn't quite land because in such an assumption, if I solve for the $+,$ then I derived that the result was still dependent on $x$, that $1 = 2x - 2a.$ If I took $-,$ then I derived $1=0.$
Pretty dismal so far, but now I'm wondering whether there exists an $f(x)$ such that
$$h(f(x)) = \frac{1}{f(x)}?$$
You can plug $f$ into $h$ and equate the two but this only finds constant solutions. How do I know there isn't some elementary or implicitly elementary or gamma-function-like integral, non-constant function that satisfies
$$\frac{1-f(x)}{f(x)^2} = \frac{1}{f(x)}?$$
$$ \begin{aligned} \frac{1-f(x)}{f(x)^2} &= \frac{1}{f(x)} \\ \frac{1}{f(x)^2}-\frac{1}{f(x)} &= \frac{1}{f(x)} \\ \frac{1}{f(x)^2} &= 2\frac{1}{f(x)} \\ f(x)^2 &= \frac{f(x)}{2} \\ f(x) &= \frac{1}{2} \end{aligned} $$