$\newcommand{\Z}{\mathbb Z}$
$\newcommand{\T}{\mathbb T}$

It is well know that the Poisson boundary of simple random walk on graph is not invariant under quasi isometries. Here's a construction:

Take $\Z^4$ and notice that (1) Its Poisson boundary is trivial (i.e. it is Liouville) and (2) a random walk starting anywhere on the line $L=\{(n,0,0,0)\mid n\in\Z \}$ has positive probability to never hit this line.

Let $\T=\{0,1\}^*$ be the infinite binary tree and let $S$ be the set of *balanced* vertices, that is, vertices with the same number of 0s and 1s in their name. Connect the vertices of $S$ with those of $L$ in an arbitrary 1-1 manner. Call the resulting graph $G$.

Now, a random walk on $G$, started anywhere in the tree, will hit $S$ almost surely. When it does, there is a positive probability for it to be absorbed in the $\Z^4$ part (i.e. to stay there forever). If this doesn't happen, then the walk will just hit $S$ again and again until it is absorbed. Since $\Z^4$ was Liouville, we get that $G$ is also Liouville.

However, consider the graph $G'$ which is identical to $G$ except that we replace each right going edge of the tree $\T$ with a path of length 2. This new graph is clearly quasi-isometric to $G$, but now there is a positive probability that the random walk will never hit $S$ and stay in the tree part of the graph forever. This give rises to a non-constant, bounded harmonic function $f(v)$ which is simply the probability of a random walk started at $v$ to be absorbed in $\Z^4$.

Compactifications of symmetric and locally symmetric spaces). So if quasi-isometry does not preserve the geodesic boundary then it does not preserve the Martin one either. Points of the geodesic boundary are equivalence classes of (unit speed, I think) geodesics $\gamma(t)$ such that $\limsup_{t\to\infty} d(\gamma_1(t),\gamma_2(t)) < +\infty$. How does a quasi-isometry act on geodesics? – Igor Khavkine Sep 13 '12 at 9:13