How would one go about generating examples of metric spaces that are uniformly isomorphic (i.e. there is a uniformly continuous bijection between them with uniformly continuous inverse), but also fail to be bi-Lipschitz equivalent (bi-Lipschitz equivalent meaning there is a Lipschitz bijection between them with a Lipschitz inverse)? Furthermore, how could this be shown from first principles i.e. definitions of Lipschitz and uniform continuity?
I can think of examples e.g. R,d and R,f(d) where d is the Euclidean norm and f is not Lipschitz (or its inverse is not Lipschitz). However, I struggle to find a way to prove these spaces are not bi-Lipschitz equivalent i.e. there is no Lipschitz bijection with a Lipschitz inverse between them.
$(\Bbb R, d_1)$ and $(\Bbb R, d_2)$ are examples when $d_1(x,y)=|x-y|$ and $d_2(x,y)=\min(1, |x-y|)$. These are uniformly equivalent (they induce the same uniformity and topology on $\Bbb R$) but there is no constant $C>0$ so that $d_1(x,y) \le Cd_2(x,y)$ for all $x,y \in \Bbb R$.