I would like to find a pair of relatively prime polynomials $p,q \in k[x]$ (where $k$ is a field) such that $$\frac{p(x)}{q(x)} = \frac{p \left( \frac{1}{1-x} \right)} {q\left( \frac{1}{1-x} \right)} .$$
I believe that $p$ and $q$ should have at most degree 3.
A similar situation is the following:
Take $f(x) = \frac{x^2+1}{x} \in k(x)$. Then
$$f(x) = f\left( \frac{1}{x} \right).$$ This problem I am currently dealing with seems much more elusive.
Let $r(x) = 1/(1-x)$. Note that $r(r(r(x))) = x$. Therefore, the rational function $$f(x) = x + r(x) + r(r(x)) = \frac{-x^3 + 3x - 1}{x(x-1)}$$ satisfies $$f\left(\frac{1}{1-x} \right) = r(x) + r(r(x)) + r(r(r(x))) = r(x) + r(r(x)) + x = f(x)$$ as desired.