Quasi-isometry group of $\{n^3;n\in\mathbb{Z}\}$. Exercise $5.E.6$ of Clara Löh's $\textit{Geometric Group Theory. An Introduction}$ asks me to determine the quasi-isometry group of $A=\{n^3 | n ∈ Z\}$. If I change that $n^3$ by, for example, $n!$, it is not hard to prove that the quasi-isometry group of $\{n! | n ∈ Z\}$ has just one element.
But the quasi-isometry group of $A$ seems much harder to find. Any solutions/hints for the problem are welcome. I leave below some of my thoughts on the problem.
Any quasi-isometries $F:A\to A$ can be expressed as $F_f:A\to A;F_f(n^3)=f(n)^3$, where $f:\mathbb{Z}\to\mathbb{Z}$ is a function such that:

*

*The image of $f$ is all $\mathbb{Z}$ except finitely many elements (because the image of $F_f$ is $M$-dense in $A$ for some constant $M$).


*$f$ is injective in $\mathbb{Z}\setminus B$, for some finite set $B$ (also a consequence of $F_f$ being a quasi-isometry).
Using this notation, two quasi-isometries $F_f$ and $F_g$ are close iff $f-g$ has finite support. My conjecture is that $f$ will be of the forms $f(n)=n+B(n)$ or $f(n)=-n+B(n)$, where $B(n)$ has to be bounded and it can be any bounded function such that $f(n)$ satisfies the two conditions above. But proving that seems like it would be really tedious, if it even is true.
 A: Yes it's true. The nontrivial input is the following result of Benjamini and Shalov: every bilipschitz permutation of $\mathbf{Z}$ is at bounded distance from either identity or minus identity. (Bilipschitz bijections of $\mathbf{Z}$, Anal. Geom. Metr. Spaces 2015; 3:313–316. link).
Now let $f:\mathbf{Z}\to\mathbf{Z}$ be as you described. One easily checks that $f$ is Lipschitz. Since $f$ is also proper, it easily follows that $f$ maps $+\infty$ to either $+\infty$ or $-\infty$. Up to compose with $n\mapsto -n$ you can suppose that $f$ maps $+\infty$ to itself. So we have to show that $f$ is at bounded distance to the identity. We know that $f$ is injective on $[m,+\infty[$. Up to postcompose with a finitely supported permutation and then a translation, we can suppose that $f$ induces a permutation of $[m,+\infty[$. Let $f'$ extend $f$ as identity outside $[m,+\infty[$. Then $f'$ is a bilipschitz permutation of $\mathbf{Z}$. By Benjamini-Shalof, $f'$ has bounded distance to identity.
So, on $\mathbf{N}$, $f$ has bounded distance to the identity. Similarly, this is true on $-\mathbf{N}$.
(Note that $f$ itself is not necessarily induced by a permutation of $\mathbf{Z}$: the obstruction is given by a homomorphism from the given group onto $\mathbf{Z}$.)
