Let $(X, \rho)$ be a compact metric space, and $T : X \to X$ a homeomorphism of $X$ to itself. I know that this $T$ can fail to be Lipschitz, e.g. $X = [0, 1], Tx = \sqrt{x}$ with the standard metric $\rho(x, y) = |x - y|$ (I found this in another post on here, but I closed the tab before I could link that post here). What's less obvious to me, however, is whether I could give $[0, 1]$ another metric $\rho'$ compatible with the standard topology on $[0, 1]$ with respect to which $\sqrt{\cdot}$ is Lipschitz.
More generally, given a compact metrizable space $X$ and a homeomorphism $T : X \to X$ of $X$ to itself, is there a way to look at the topological properties of $T$ and discern whether $T$ could be made Lipschitz by endowing $X$ with an appropriate metric? I know there are several metrical properties which can be dependent upon which metric is being used, e.g. boundedness and Cauchy-ness, but I don't know to what extent Lipschitzness is visible in the topology.
For any continuous $T\colon X\to X$ (not necessarily a homeomorphism) and any metric $\rho$ on $X$ (with finite diameter, which is automatic if $X$ is compact), you can define a new metric $\rho'\colon X\times X\to[0,2\operatorname{diam}(X,\rho)]\subset\mathbb{R}$ by $$ \rho'(x,x'):=\sum_{n\geq 0}2^{-n}\rho(T^nx,T^nx') $$ Then $\operatorname{Lip}_{\rho'}(T)\leq 2$: \begin{align*} \rho'(Tx,Tx')&=\sum_{n\geq 0}2^{-n}\rho(T^{n+1}x,T^{n+1}x')\\ &=2\sum_{n\geq 1}2^{-n}\rho(T^nx,T^nx')\\ &\leq2\sum_{n\geq 0}2^{-n}\rho(T^nx,T^nx')=2\rho'(x,x') \end{align*}