What general techniques are there to show that two finitely-generated groups are bi-Lipschitz equivalent? We define bi-Lipschitz equivalency as follows.

We say two metric spaces $(X,d)$, $(Y,e)$ are bi-Lipschitz
equivalent if there exists a bijection $f : X \to Y$ and some $ L
 \geq 1$ such that  $$ \frac 1 L \ d(x,y) \leq e(f(x),f(y)) \leq L \
 d(x,y), $$ for all $x,y \in X$.

Viewing finitely-generated (f.g.) groups as metric spaces under their word metric, we have that any two word metrics on a f.g. group are bi-Lipschitz equivalent. Thus it makes sense to discuss whether two f.g. groups $G$, $H$ are bi-Lipschitz equivalent, without fixing any generating sets.
I understand that to show two groups are not bi-Lipschitz,  we may compare their growth functions, for example. However I am not aware of any general tools for showing two groups are bi-Lipschitz equivalent, outside of just exhibiting such a map $f$ as above (and even that seems incredibly difficult).
Any references to literature which discusses this topic or contains examples would be appreciated. Thanks!
 A: Whyte, Kevin, Amenability, bi-Lipschitz equivalence, and the von Neumann conjecture, Duke Math. J. 99, No. 1, 93-112 (1999). ZBL1017.54017.
A: I'd say:

*

*For non-amenable groups, the theorem by Whyte mentioned by Moishe without comment is that every quasi-isometry is at bounded distance from a bilipschitz map. Hence the question boils down to the same with quasi-isometry, which is broad and widely studied. Among others, all cocompact lattices in the same locally compact group are quasi-isometric.


*For amenable groups, this is trivially false (finite groups of distinct cardinal are QI but not bilipschitz), Much harder, Dymarz exhibited (solvable) infinite examples. So I guess there are few general methods. It is enough to have bilipschitz subgroups of the same finite index. For instance, $\mathbf{Z}$ and $D_\infty$ are bilipschitz since they both have an index 2 copy of $\mathbf{Z}$. Also, some constructions pass to bilipschitz maps (rather than QI): if $A,B$ are bilipschitz, then $A\wr\Gamma$ and $B\wr\Gamma$ are bilipschitz (the analogous statement with QI being false), the proof being the obvious naive one.
There are many open questions, for instance it's not known whether there exists two QI but not bilipschitz infinite f.g. nilpotent groups. It's not known if there exists a f.g. nilpotent group that is  not bilipschitz to one of its finite index subgroups.
