Given a metric space $(X, d)$, we can consider the set of all quasi-isometries $f: X \to X$, and quotient out by the equivalence relation identifying $f$ and $g$ if $\sup\_{x \in X}d(f(x), g(x))$ is finite. Doing so, we obtain a set of equivalence classes $\mathcal{QI}(X)$ that is a group under composition.


In the same spirit as the questions [Every group is a fundamental group](https://math.stackexchange.com/questions/939856/every-group-is-a-fundamental-group) and [Is every group the automorphism group of a group?](https://math.stackexchange.com/questions/253936/is-every-group-the-automorphism-group-of-a-group), we can ask: for which groups $G$ does there exist a metric space $X$ such that $\mathcal{QI}(X) \cong G?$


Surprisingly, someone told me today that basically nothing is known about this question. According to them, we do not even know how to construct a metric space $X$ such that $\mathcal{QI}(X)$ is a finite cyclic group.


Given this, my question is: what *do* we know about the quasi-isometry groups of metric spaces? For example, what are some metric spaces $X$ for which $\mathcal{QI}(X)$ has been computed? Do we know of any groups $G$ which are not isomorphic to $\mathcal{QI}(X)$ for any $X$?



Quasi-isometry groups of metric spaces
A first observation is that $\mathcal{QI}$ is quite complicated for most natural spaces. For instance, any two linear maps $x \mapsto \lambda x, x \mapsto \lambda' x$ are equal in $\mathcal{QI}(\mathbb{R})$ if and only if $\lambda = \lambda'$. A corollary of this is that $\mathcal{QI}(\mathbb{N})$ is uncountable.


In order to get small $\mathcal{QI}(X)$ to be small, one must spread apart points in $X$ to sabotage the flexibility of the quasi-isometry condition. A helpful building block and motivating example is $X\_0 = \{ n! \mid n \in \mathbb{N} \} \subset \mathbb{N}$. Any $(K,C)$-quasi isometry $f$ of $X\_0$ must have $f(n!) = n!$ for all $n$ large enough, for instance $n > \text{max}(K, C) + 1$, by considering $d(f(n!), f((n+1)!))$. So $\mathcal{QI}(X\_0)$ is trivial. If one allows metrics that obtain $\infty$ as a distance, one obtains a space with $\mathcal{QI}(X\_n) = S\_n$ (with $S\_n$ the symmetric group on $n$ letters) by taking $n$ copies of $X\_0$, with infinite distance between any two copies.


An addendum: this construction also gives direct products of symmetric groups by mixing growth rates in building blocks. For instance, let $Y\_0 = \{ (n!)! \, | n \in \mathbb{N}\}$. Then taking the disjoint union of $n$ copies of $X\_0$ and $m$ copies of $Y\_0$ as above gives a space with $\mathcal{QI}(X) = S\_n \times S\_m$. This is because for a $(K, C)$ quasi-isometry, one can not map sufficiently large elements of $X\_0$ into $Y\_0$ or vice-versa. One sees this by comparing the distances between $3$ consecutive elements in $X\_0$ or $Y\_0$.


A correction: as pointed out by Fedya below, a previous version of this answer incorrectly asserted that spaces $X\_k$ obtained as pinwheels of $k$ copies of $X\_0$ have $\mathcal{QI}(X\_k) \cong S\_k$. One can independently permute each of the $k$ copies of $n!$ while remaining a quasi-isometry, and finite-distance maps allow one to freely move around finitely many points freely. This yields the quasi-isometry group of $X\_k$ to be isomorphic to $(\prod\_{k=1}^{\infty}S\_k) / (\bigoplus\_{k=1}^\infty S\_k).$


It seems quite unclear how to build many other groups with explicit examples. In particular, one question I think is interesting but do not know how to answer is if there exist metric spaces (with only finite distances allowed) with finite and nontrivial $\mathcal{QI}$ group.


