Quasi-isometry groups of metric spaces 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 and 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 $X$?
 A: Regarding the question of spaces $X$ for which $QI(X)$ is known, a good keyword is quasi-isometric rigidity. One reference would be the survey in Chapter 25 of the Druţu–Kapovich book "Geometric group theory".
A space $X$ is called strongly QI rigid if the map $\operatorname{Isom}(X) \to QI(X)$ is surjective. Often $X$ has no non-trivial isometries that are bounded distance from the identity, so we get an isomorphism. For example, Pansu showed that quaternionic hyperbolic spaces are strongly QI rigid.
Pansu, Pierre, Carnot-Carathéodory metrics and quasiisometries of symmetric spaces of rank 1, Ann. Math. (2) 129, No. 1, 1-60 (1989). ZBL0678.53042..
A: 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.
