# Id $f,g$ orientation preserving cricle-diffeomorphisms, then $\rho(g^{-1}\circ f \circ g) = \rho(f)$.

Let $$F, G : \mathbb{R} \rightarrow \mathbb{R}$$ be a lift of $$f$$ and $$G$$ of $$g$$. That is, $$\pi \circ F = f \circ \pi$$ with $$\pi(x) = e^{2\pi i}$$. We define $$\rho_{0}(F) = \lim_{n\rightarrow \infty}\frac{F^{n}}{n}.$$ One property: for $$F,F'$$ two different lifts of $$f$$, $$|\rho_{0}(F) - \rho_{0}(F)| \in \mathbb{Z}.$$ And another: $$G^{-1}\circ F \circ G$$ is a lift for $$g^{-1}\circ f \circ g$$. The rotation number of $$f$$, $$\rho(f)$$, is the fraction part of $$\rho_{0}(F)$$ for any lift $$F$$ of $$f$$. That is, $$\rho(f)$$ is the unique number in $$[0,1)$$ such that $$\rho_{0}(F) - \rho(f)$$ is an integer.

Can you help me prove the title-statement?

I got the hint that $$\lim_{n\rightarrow\infty} \frac{(G^{-1} \circ F \circ G)^{n}(x)}{n} = \lim_{n\rightarrow\infty}\frac{G^{-1}\circ F^{n} \circ G(x)}{n}$$ and also that when this last expression equals $$\rho_{0}(F)$$ we have that the requested result follows. But how to finish this?