Statement: Let $(X,d)$, $(Y,r)$ be homeomorphic compact metric spaces. Then, there exists a bilipschitz homeomorphism between the two.
Problem: As I have no clue whether the statement is true or false (I'm not well acquainted with the theory of lipschitz functions), any proof/counterexample is welcome.
Rephrasing: Define the Lipschitz distance of two metric spaces as $$d_L(X,Y)\doteq \inf\big|\ln \mathrm{dil}(f)+\ln \mathrm{dil}(f^{-1})\big|$$ where $\mathrm{dil}(f)$ is the dilatation of $f$ and the infimum is taken over all bilipschitz homeomorphisms between X and Y. The question is now whether the Lipschitz distance between two homeomorphic compact metric spaces is always finite.
Comments: It is clear to me that not all homeomorphisms between compact metric spaces are bilipschitz: $x^3\colon [-1,1]\rightarrow [-1,1]$ is a counterexample, which, although trivial, suggests that simply scaling the metrics won't do the trick. Also, any homeomorphism between compact metric spaces is uniformly bicontinuous, thus any such spaces are uniformly isomorphic (where, say à la Weil, uniformities are induced by the respective metrics), which is encouraging towards an affirmative answer, yet it seems difficult to prove (disprove) the existence of any bilipschitz homeomorphism given that the spaces are effectively homeomorphic.
As I would like the statement to be true, I was wondering the following: every compact metric space is complete and second countable ($\Rightarrow$ separable), i.e. Polish, so it is embeddable in Tikhonov's cube $T\doteq I^\mathbb N$. Is there any property of $\mathrm{Homeo}(T)$ that could be of help here?