Let $\Gamma\subset \text{Isom}(\mathbb{H}^n)$ be a discrete group of isometries. Suppose that $\Gamma$ fixes a finite set of points in $\overline{\mathbb{H}^n}$ (as a set, not one by one). Let $\phi_1\in \Gamma$ be an hyperbolic isometry of $\mathbb{H}^n$ fixing the points $\{0,\infty\}$ (seeing $\mathbb{H}^n$ with the half plane model). Let also $\phi_2\in \Gamma$ be an elliptic isometry fixing the point $\infty$. Is it true that $\phi_2$ must either fix $0$ or fix the horospheres centered at $\infty$?
Here's what i now: $\phi_2$ must of course permute the horospheres centered at $\infty$. Suppose that the permutation is non trivial and that $\phi_2$ sends the horosphere $O_0$ at height $t_0$ to the one $O_1$ at height $t_1$ (in the half space model). WLOG $t_1<t_0$. In $O_0$ the distances are shorter, so the map $\psi:O_1\to O_0$ sending $(x,t_1)$ to $(x,t_0)$ is a contraction. Then $\phi_2\circ \psi:O_1\to O_1$ is a contraction and has a fixed point $(y,t_1)$. Then $\phi_2(y,t_0)=(y,t_1)$ and $\phi_2$ fixes the geodesic passing through $\infty$ and $(y,t_0)$. This means that $\phi_2$ also fixes $(y,0)$.
My question is: must $(y,0)=(0,0)$ under the initial hypothesis?
Let $A \subset \overline{\mathbb H}$ be the finite set that is fixed by $\Gamma$. Your post does not say, but I will presume that $A$ is not empty.
Since $\gamma \in \Gamma$ has infinite order, the group $\Gamma$ is infinite. We may conclude that $A$ is a subset of the sphere at infinity $\partial \mathbb H = \overline{\mathbb H} - \mathbb H$ containing exactly 1 or 2 points, because: a discrete group that preserves a nonempty finite subset of $\mathbb H$ is finite; and a discrete group $\Gamma$ that preserves a set $A \subset \partial\mathbb H$ consisting of 3 or more points also preserves a nonempty finite subset of $\mathbb H$, namely the "barycenters" of each three-point subset of $A$, and so $\Gamma$ is finite.
If $A$ contains exactly two points then it follows, from your assumptions on $\phi_1$, that $A = \{0,\infty\}$, and then it follows that $\phi_2$ also preserves $A$. Since $\phi_2$ fixes $\infty$ it follows that $\phi_2$ fixes $0$.
If $A$ contains exactly one point then, again from your assumptions on $\phi_1$, that point is either $0$ or $\infty$. If that point is $0$ then $\phi_2$ also fixes $0$.
So we are reduced to the case $A = \{\infty\}$. Thus the entire group $\Gamma$ preserves the horospheres centered at $\infty$, and so $\phi_2$ also preserves the horospheres centered at $\infty$.
So, in every case we have proved that $\phi_2$ either fixes $0$ or preserves the horospheres centered at $\infty$ and the proof is done. I notice that we have not used the hypothesis that $\phi_2$ is elliptic.