Take two graphs (of bounded valency) or manifolds (f bounded geometry) $G$ and $G'$. Assume there is a quasi-isometry $f:G \to G'$, and assume the Poisson or Martin boundary of $G$ is known, what may one say about the Poisson or Martin boundary of $G'$ (or how does $f$ transforms Poisson or Martin boundaries)? 


EDIT: a map $f:X \to Y$ between two metric spaces $(X,d\_X)$ and $(Y,d\_Y)$ is a quasi-isometry if $\exists C \geq 0$ and $D \geq 1$ such that


* $\forall x,x' \in X$, $D^{-1} d\_Y \big( f(x), f(x') \big) - C \leq d\_X(x,x') \leq D d\_Y \big( f(x), f(x') \big) + C $
* $\forall y \in Y, d\_Y \big(f(X),y \big) < C$



How does a quasi-isometry affect Poisson or Martin boundaries?
1. Terry Lyons in "Instability of the Liouville property for quasi-isometric Riemannian manifolds and reversible Markov chains", Journal of Diffe Geometry, 1987, constructs examples of graphs (and Riemannian manifolds) $G, G'$ which are quasi-isometric, but one admits no (nonconstant) positive harmonic functions while the other admits bounded (nonconstant) harmonic functions. Thus, quasi-isometries cannot respect Martin (and Poisson, as Ori noted) 
boundaries in any meaningful way.
2. It is well-known that, in general, even for CAT(0) spaces, quasi-isometries do not induce maps of geometric boundaries. The easiest example to consider is the case of quasi-isometries of the Euclidean plane ${\mathbb R}^2$. Let $f(x)$ be any $L$-Lipschitz function ($L\ge 1$) of one variable (you can take, for instance, $f(x)=|x|$). Define a self-map $F(x,y)= (x, f(x)+y)$ of ${\mathbb R}^2$. Then $F$ is an $2L$-bilipschitz homeomorphism of ${\mathbb R}^2$. Considering images of the $x$-axis under $F$, we see that they are (in general) not close to geodesics (say, the graph of $y=|x|$). Moreover, Croke and Kleiner constructed example ("Spaces with nonpositive curvature and their ideal boundaries", Topology, 2000) of a CAT(0) group $G$ so that $G$ acts geometrically (discretely, isometrically, cocompactly) on two CAT(0) spaces $X, X'$ so that the geometric boundaries of $X, X'$ are not homeomorphic. Existence of geometric action of $G$ on $X, X'$ implies that these spaces are quasi-isometric to each other.
3. Still, for some interesting classes of spaces, quasi-isometries act naturally on all three boundaries (geometric, Poisson, Martin); this happens for instance, in the case of Cayley graphs of hyperbolic groups.


