Let $X,Y$ be metric spaces and suppose there exist bijective Lipschitz functions $f : X \rightarrow Y$ and $g : Y \rightarrow X$. Does there necessarily exist a bijective bi-Lipschitz function $h : X \rightarrow Y$?
Thinking about it for a while, I couldn't see any obvious path to proving this, but coming up with a counter-example seems very difficult as well. I couldn't find it in any reference, which leads me to believe it's probably false, since a result like this would probably be pretty well-known. I'm also interested in knowing if this might be true with some additional restrictions.
Here's a counterexample.
For the space $X$, take the disjoint union of intervals $$[0,1], [0,2], [0,3], [0,4], [0,5], [0,6], .... $$ and identify their initial $0$ endpoints to get the metric space $X$. The distance between two points $x,y$ on the same interval $[0,n]$ is $|x-y|$, whereas the distance between two points $x \in [0,m]$, $y \in [0,n]$ on different intervals, with $m \ne n$, is $x+y$.
For the space $Y$, do the same thing but with the disjoint union of intervals $$[0,1], [0,2], [0,4], [0,8], [0,16], [0,32], ... $$ Map $X$ to $Y$ by including $[0,n]$ into $[0,2^n]$ isometrically. Map $Y$ to $X$ by identifying $[0,2^n]$ with $[0,2^n]$ isometrically. Each of these is actually an isometric embedding, i.e. a bi-Lipschitz embedding with bi-Lipschitz constant $1$.
But for any $L \ge 1$ there is no $L$ bi-Lipschitz bijection $f : X \to Y$, because $f$ would have to map the $0$ point of $X$ to the $0$ point of $Y$, and there would have to be a bijective enumeration $k_n$ of the natural numbers such that $f[0,n] = [0,2^{k_n}]$. It would then follow (by the pigeonhole principle) that there exist arbitrarily large $n$ such that $k_n \ge n$, and a contradiction ensues by taking such an $n$ so large that $2^{k_n} \ge 2^n / n > L$, hence there is no $L$-bilipschitz bijection between $[0,n]$ and $[0,2^{k_n}]$.