Liouville's Theorem for Conformal Maps states that every conformal map defined on a domain in $\mathbb{R}^n$ ($n\geq 3$) is the composition of translations, dilations, rotations, and inversions through spheres. Using this theorem, it is possible to prove that every conformal automorphism of $H^n$ is a hyperbolic isometry.
It follows from this that $\mathrm{Aut}(H^n)$ is isomorphic to the group $SO^+(1,n)$ (an indefinite special orthogonal group). Thus, one way of keeping track of such automorphisms is to use $(n+1)\times(n+1)$ matrices of a certain type. In particular, let $\mathrm{Hyp}$ denote the hyperboloid $-x_0^2 + x_1^2 + \cdots + x_n^2 = -1$ in $\mathbb{R}^{n+1}$, and let $\mathrm{Hyp}^+$ be the portion of the hyperboloid for which $x_0>0$. Then $SO^+(1,n)$ acts on $\mathrm{Hyp}^+$ via linear transformations. If we conjugate by the map $f\colon \mathrm{Hyp}^+\to H^n$ defined by
$$
f(x_0,x_1,\ldots,x_n) \;=\; \left(\frac{x_2}{x_0+x_1},\ldots,\frac{x_n}{x_0+x_1},\frac{1}{x_0+x_1}\right)
$$
then we get an action of $SO^+(1,n)$ on $H^n$.
Unfortunately, this description doesn't give much geometric insight, and isn't closely related to the upper half-space $H^n$. What follows is a more geometric description of these automorphisms. I'm not sure whether or not it is helpful, but it's much more closely related to the upper half-space model. At the very least, it gives you a good way of constructing examples of conformal automorphisms of $H^n$.
Consider a hyperbolic isometry $\alpha\colon H^n\to H^n$. As in dimensions two and three, any such isometry extends to the hyperbolic boundary, which is $\mathbb{R}^{n-1}\cup\{\infty\}$. There are three cases:
Case 1: The isometry $\alpha$ fixes $\infty$. In this case, $\alpha$ must simply be a Euclidean similarity $H^n\to H^n$. In particular, $\alpha$ has the form
$$
\alpha(\textbf{x},z) = (kA\textbf{x}+\textbf{b},kz)
$$
for all $(\textbf{x},z) \in\mathbb{R}^{n-1}\times(0,\infty)$, where $k$ is a positive real number, $\textbf{b}\in\mathbb{R}^{n-1}$, and $A\in SO(n-1)$. (Note that any map of this form is indeed a conformal automorphism of $H^n$.)
Case 2: The isometry $\alpha$ maps the origin to $\infty$. In this case, let $\rho\colon H^n\to H^n$ be the map
$$
\rho(\textbf{x},z)=\frac{1}{\|\textbf{x}\|^2 + z^2}(B\textbf{x},z)
$$
where $B$ is some orientation-reversing element of $O(n-1)$, e.g. negation of the first coordinate. (The map $\rho$ is essentially the composition of inversion through the unit sphere with the reflection $B$.) Then $\rho$ is an automorphism of $H^n$ that maps the origin to $\infty$, and $\alpha$ can be expressed as $\beta\circ \rho$, where $\beta$ is some automorphism of $H^n$ that fixes $\infty$ (see case 1).
Case 3: The isometry $\alpha$ maps some other boundary point $(\textbf{p},0)$ to $\infty$. In this case, let $\tau$ be the translation $\tau(\textbf{x},z) = (\textbf{x}-\textbf{p},z)$. Then $\alpha$ can be written as $\beta\circ\tau$, where $\beta$ is an automorphism of $H^n$ that maps the origin to $\infty$.
Conlusion: Every conformal automorphisms of $H^n$ is either a Euclidean similarity transformation, or can be expressed uniquely as the composition of a translation $\tau$, the map $\rho$, and a Euclidean similarity transformation.
