Cantor-Bernstein for quasi-isometric embeddings? Suppose that two finitely generated groups quasi-isometrically embed into each other. Does it follow that the two groups are quasi-isometric? Recall that a quasi-isometry is a quasi-isometric embedding that is quasi-surjective, see e.g. https://www.math.ucdavis.edu/~kapovich/EPR/pc_lectures3.pdf
 A: 1) LAMPLIGHTER GROUPS
As mentioned by Yves, lamplighter groups over $\mathbb{Z}$ provide counterexamples thanks to Eskin, Fisher, and Whyte's work. Other counterexamples are given by lamplighter groups over one-ended groups thanks to the recent preprint arxiv:2105.04878. More precisely, if $F_1,F_2$ are two finite groups and if $H$ is a finitely presented one-ended group, then $F_1\wr H$ and $F_2 \wr H$ quasi-isometrically embed into each other, but they are not quasi-isometric if $|F_1|,|F_2|$ do not have the same prime divisors. (The converse holds if $H$ is non-amenable. If $H$ is amenable, $|F_1|,|F_2|$ must be powers of a common number and $H$ must admit specific auto-quasi-isometries (which always the case if $H$ is free abelian).) The advantage of these lampligthers is that one can avoid the heavy machinery of coarse differentiation.
2) COXETER GROUPS
Counterexamples can also be found among Coxeter groups. I just give one example, based on the following statement:
Proposition: Let $\Gamma$ be a finite simplicial graph and $u \in V(\Gamma)$ a vertex. Let $\Gamma(u)$ denote the graph obtained by gluing two copies of $\Gamma \backslash \{u\}$ along $\mathrm{link}(u)$. Then the right-angled Coxeter groups $C(\Gamma)$ and $C(\Gamma(u))$ are commensurable.
Sketch of proof. Let $u_1,\ldots, u_k$ denote the vertices of $\Gamma \backslash \mathrm{star}(u)$. Then, by a standard ping-pong argument, one proves that the reflexion subgroup $\langle u_1, \ldots, u_k,u_1^u, \ldots, u_k^u \rangle$ admits $\{u_1, \ldots, u_k,u_1^u, \ldots, u_k^u\}$ as a basis. The commutation graph of this family is precisely $\Gamma(u)$. $\square$
(Some details can be found in arxiv:1910.04230; see Proposition 3.14.)
Now, we apply the proposition to the following graph:

Clearly, $\Gamma(u)$ contains an induced copy of $\Gamma \sqcup \{ \mathrm{pt} \}$, so $C(\Gamma)$ quasi-isometrically embeds into the free product $C(\Gamma) \ast \mathbb{Z}/2\mathbb{Z}$. The reverse quasi-isometric embedding is clear. But $C(\Gamma)$ is one-ended, because $\Gamma$ has no separating complete subgraph, so $C(\Gamma)$ and $C(\Gamma) \ast \mathbb{Z}/2\mathbb{Z}$ cannot be quasi-isometric.
3) RIGHT-ANGLED ARTIN GROUPS
Given a simplicial graph $\Gamma$, the associated right-angled Artin group is defined by the following presentation:
$$A_\Gamma = \langle V(\Gamma) \mid [u,v]=1, \ \{ u,v \} \in E(\Gamma) \rangle$$
where $V(\Gamma)$ and $E(\Gamma)$ denote the vertex- and edge-sets of $\Gamma$ respectively.
First, a general statement. A graph $\Gamma$ is join if there exists a partition $V(\Gamma)= A \sqcup B$ such that $A,B$ are both non-empty and such that any vertex of $A$ is adjacent to any vertex of $B$.
Proposition: Let $\Gamma$ be a finite simplicial graph which is not a join and which is not reduced to a single vertex. Then $A_\Gamma$ contains a quasi-isometrically embedded subgroup isomorphic to the free product $A_\Gamma \ast \mathbb{Z}$.
Sketch of proof. Let $\Gamma^e$ denote the extension graph of $\Gamma$, ie., the graph whose vertices are the conjugates $gug^{-1}$, where $g \in A_\Gamma$ and $u \in V(\Gamma)$, and whose edges link two elements when they commute. In Embedability between right-angled Artin groups, Kim and Koberda prove that, if $\Lambda$ is a finite induced subgraph of $\Gamma^e$, then $A_\Lambda$ embeds into $A_\Gamma$. It turns out that $\Gamma^e$ is unbounded since $\Gamma$ is not a join, so that $\Gamma^e$ contains an induced subgraph isomorphic to $\Gamma \sqcup \{ \text{vertex} \}$, hence $A_\Gamma \ast \mathbb{Z} \leq A_\Gamma$.
I don't know if it follows from their argument that the embedding is quasi-isometric, but, in the alternative proof I gave of their theorem in my thesis (see Section 8.5), it is not difficult to show that the embedding I construct is quasi-isometric. $\square$
Corollary: If $\Gamma$ is a finite connected simplicial graph which is not a join, then $A_\Gamma$ and $A_\Gamma \ast \mathbb{Z}$ quasi-isometrically embeds into each other, but they are not quasi-isometric.
The fact that $A_\Gamma$ and $A_\Gamma \ast \mathbb{Z}$ are not quasi-isometric follows from the observation that $A_\Gamma$ is one-ended (as $\Gamma$ is connected and contains at least two vertices).
Now, a concrete example. Consider the right-angled Artin group
$$A= \langle a,b,c,d \mid [a,b]=[b,c]=[c,d]=1 \rangle.$$
Of course, $A$ quasi-isometrically embeds into $A \ast \mathbb{Z}$. Conversely, an embedding as mentioned above shows that the subgroup
$$\langle a,b,c,d^2,ada^{-1} \rangle= \langle a,b,c,d^2 \rangle \ast \langle ada^{-1} \rangle \simeq A \ast \mathbb{Z}$$
is quasi-isometrically embedded into $A$, which can also be proved directly. However, $A$ and $A \ast \mathbb{Z}$ are not quasi-isometric since $A$ is one-ended.
4) A HYPERBOLIC EXAMPLE
According to arxiv:1812.07799, there exist a torsion-free one-ended hyperbolic group $G$ containing two isomorphic quasiconvex subgroups $H_1,H_2 \leq G$ such that $H_1$ (resp. $H_2$) has finite (resp. infinite) index in $G$. As a consequence of a theorem claimed by Gromov (and proved by Arzhantseva), $G$ contains a subgroup isomorphic to the free product $H_2 \ast \mathbb{Z}$. As a consequence, $G$ and $G\ast \mathbb{Z}$ quasi-isometrically embed into each other but they are not quasi-isometric.
A: Let $C_n$ be a cyclic groups of order $n$. Then the wreath products $C_2\wr\mathbf{Z}$ and $C_3\wr\mathbf{Z}$ embed QI into each other (for the reverse direction, observe that $C_2\wr\mathbf{Z}$ has a subgroup of index 2 isomorphic to $C_2^2\wr\mathbf{Z}$). But $C_2\wr\mathbf{Z}$ and $C_3\wr\mathbf{Z}$ are not QI, this follows from work of Eskin-Fisher-Whyte ($F\wr\mathbf{Z}$ and $F'\wr\mathbf{Z}$ for $F,F'$ finite groups are QI iff $|F|,|F'|$ have some common power.)
