Let $X$ be a rank $1$ symmetric space of non-compact type and $G$ its isometry group. $G$ is a semisimple linear algebraic Lie group of non-compact type with trivial center. Let $\rho$ be a representation of a discrete group $\Gamma$ into $G$. It is known that if the Zariski closure of $\rho$ is reductive then there is a finite subset $F\subset \Gamma$ and a constant $D\geq 0$ such that for any $\gamma \in \Gamma$ there is an $f\in F$ such that $$|\mu(\rho(\gamma f)) - \lambda(\rho(\gamma(f))|\leq D$$ Here, $\mu$ is the displacement function with respect to a fix point $p$, $$\mu(g) = d(p,g\cdot p)$$ and $\lambda$ is the translation length $$\lambda(g) = \inf_{x\in X} d(x,g\cdot x)$$

In the literature people say that $\gamma f$ is $\textit{proximal}$. This fact is stated on page 71 of this paper: https://arxiv.org/abs/1307.0250 and the ingredients of a proof are outlined on pages 7 and 8 of this document: https://arxiv.org/pdf/0704.3499.pdf

I am wondering about the following: suppose $X$ is instead a complete $CAT(-1)$ metric space and $\rho$ is reductive, which in this case means the number of fixed points of $\rho(\Gamma)$ on the ideal boundary $\partial_\infty X$ is different than $1$ (this is equivalent to Zariski closure being reductive in symmetric space setting), does the existence of $F$ and $D$ as above still hold? This seems to me like a general fact that should not rely on the fact that we have a matrix Lie group, although the proof heavily uses this machinery.

I must admit I have not spent too long trying to prove this; if the result is already out there I don't want to reinvent the wheel. I should also comment that all I really need for my purposes is for $X$ to be a $CAT(-1)$ Hadamard manifold, but I would expect the general result to be true.