Isometry in hyperbolic space determined by induced conformal homeomorphsm on the boundary? The proposition in the book of Introduction to Geometric Topology[Martelli] states that

technically, it didn't state there is a one-to-one correspondence between the isometry inside and conformal map on the boundary.
In the two dimension boundary case, the general conformal group is much larger than the other dimensions, because each holomorphic function $z\to f(z)$ is conformal. I wonder what a general holomorphic map (which is conformal) will induce in the $\mathbb{H}^3$.
 A: You write In the two dimension boundary case, the general conformal group is much larger than the other dimensions, because each holomorphic function $z\to f(z)$ is conformal.
But this statement is wrong, because you are ignoring two important facts: not every holomorphic map is conformal; and not every conformal map has domain $\mathbb C^* = \overline{\mathbb H}^3$. These properties are necessary in order for a conformal map to be the trace of an isometry of $\mathbb H^3$; furthermore, they are sufficient. (And besides that, you need these conditions simply to ensure that you get an actual group: you need these two conditions to ensure the existence of an inverse.)
In more detail, let's consider a general holomorphic map $f : U \to \mathbb C^* = \overline{\mathbb H}^3$, where $U \subset \mathbb C^*$ is open.
First, in order for $f$ to be conformal, it is necessary (and sufficient) that $f$ have nonzero derivative at every point of its domain $U$. So, for example, $f(z)=z^2$ is defined on $U=\mathbb C \cup \{\infty\} = \overline{\mathbb H}^3$. However, $f'(0)=0$ (also, under the coordinate change $w=\frac{1}{z}$, the derivative at $\infty$ is also zero). Therefore $f'$ is not conformal, and $f'$ is not the trace of any isometry of $\mathbb H^3$.
Second, assuming now that $f$ is conformal, nonetheless in order for $f$ to be the trace of an isometry of $\mathbb H^3$, it is necessary that $U = \mathbb C^*$. For example, we can restrict $f(z)=z^2$ to the half plane $\text{Re}(z)>0$, and that restricted map is now conformal, however the domain is not all of $\mathbb C^*$ and so once again $f$ is not the trace of an isometry of $\mathbb H^3$.

So, keeping in mind those two necessary conditions, it turns out that they are also sufficient, and your statement that the general conformal group is much larger than the other dimensions is unfounded.
Here's what's true:

Theorem: If $f : \mathbb C^* \to \mathbb C^*$ is conformal then $f$ is the trace of an isometry of $\mathbb H^3$.

This is proved in a few steps.
Step 1: One considers the special case that $f(0)=0$ and $f(\infty)=\infty$, and proves that there is a complex constant $C \ne 0$ such that $f(z)=Cz$ (I think this is some version of the Schwarz Lemma but I'm not so good on names of theorems).
Step 2: One proves in the general case that $f$ is a fractional linear transformation, meaning there are complex constants $a,b,c,d$ such that $ad-bc=1$ and such that $f(z) = \frac{az+b}{cz+d}$.
Step 3: One proves that every fractional linear transformation is indeed the trace of an isometry of $\mathbb H^3$.

To summarize: the conformal group of $\overline{\mathbb H}^3 = \mathbb C^*$ is the group of fractional linear transformations of $\mathbb C^*$, and it is indeed the group of traces of isometries of $\mathbb H^3$.
