Let $K$ be a compact metrizable space. A homeomorphism $f\colon K\to K$ is called equicontinuous with respect to a (compatible) metric $d$ on $K$ if for all $\epsilon>0$ there is a $\delta>0$ such that for all $x,y\in K$ with $d(x,y)<\delta$ we have $d(f^{n}(x),f^{n}(y))<\epsilon$ for all $n\in\mathbb{Z}$.
I am trying to prove that this definition does not depend on the metric $d$. In other words, if $f$ is an equicontinuous homeomorphism with respect to $d$ and $d'$ is an equivalent metric on $K$, then $f$ is equicontinuous with respect to $d'$.
I tried to work with $\rho(x,y):=\sup_{n\in\mathbb{Z}}d(f^{n}(x),f^{n}(y))$, but it didn't really get me anywhere. Any suggestions would be greatly appreciated!
I think I managed to prove it. The key ingredient is that $\text{id}\colon(K,d)\to(K,d')$ and its inverse are uniformly continuous by compactness. Indeed, let $\epsilon>0$ be given. By uniform continuity there is a $\delta_{1}>0$ such that $d(x,y)<\delta_{1}$ implies $d'(x,y)<\epsilon$. By hypothesis there is a $\delta_{2}>0$ such that $d(x,y)<\delta_{2}$ implies $d(f^{n}(x),f^{n}(y))<\delta_{1}$ for all $n\in\mathbb{Z}$. Again by uniform continuity there is a $\delta_{3}>0$ such that $d'(x,y)<\delta_{3}$ implies $d(x,y)<\delta_{2}$. Putting everything together yields $d'(x,y)<\delta_{3}$ implies $d'(f^{n}(x),f^{n}(y))<\epsilon$ for all $n\in\mathbb{Z}$.