You could explore this conjecture by the following method: Suppose $f(f(k))= h(f(k))$ for different specified $h$, then look for $G(z) = H(F(z))$. So eg. $h(k) = a k + b$ gives $G(z) = a F(z) +\frac{b}{1-z}.$ Then you need to solve the functional equation $f(f(k)) = h(f(k))$, and this will give you some sets of pairs $(h, H)$ which might inform your conjecture

You could explore this conjecture by the following method: Suppose $f(f(k))= h(f(k))$ for different specified $h$, then look for $G(z) = H(F(z))$. So eg. $h(k) = a k + b$ gives $G(z) = a F(z) +\frac{b}{1-z}.$ Then you need to solve the functional equation $f(f(k)) = h(f(k))$, and this will give you some sets of pairs $(h, H)$ which might inform your conjecture

The example that you give in the comments on the other question is $(h,H)=(1,1)$, but this case is very simple, I don't know that I would take it to be the basis for a conjecture.