I am trying to understand why Mostow rigidity fails in dimension 2. More concretely, I have the following question:

(1) What is an example of a quasiisometry $f$ of the hyperbolic plane $\mathbb H^2$ to itself such that for the continuation $\partial f$ of $f$ to the boundary $\partial \mathbb H^2$, there does not exist an isometry $\varphi$ of $\mathbb H$ such that $\bar{\varphi} |_{\partial \mathbb H^2} = \partial f$?

The reason I am asking this question is that one way to prove classical Mostow rigidity (two closed hyperbolic manifolds of the same dimension $n \geq 3$ with isomorphic fundamental groups are isomorphic) can be deduced from the statement that for $n \geq 3$, every quasiisometry $f$ of hyperbolic space $\mathbb H^n$ extends to a homeomorphism $\partial f: \partial \mathbb H^n \rightarrow \partial \mathbb H^n$ which is the restriction of an isometry of $\mathbb H^n$.

The proof of this statement decomposes into two pieces: First, one shows that $\partial f$ is a homeomorphism, secondly, one proves that it is conformal and hence has to come from an isometry of $\mathbb H^n$. The first step also works in dimension 2, but the second does not. So more general than the above, we could even ask the following (by identifying $\partial \mathbb H^2$ with $S^1$):

(2) Which self-homeomorphisms of $S^1$ arise as extensions of quasiisometries $\mathbb H^2 \rightarrow \mathbb H^2$ to the boundary?