First, recall that a regular tree $\mathcal{T}_n$ ($n>1$), is an infinite tree in which every vertex has valency $n$. It is known that the automorphisms of a regular tree can be classified into three different types, namely
- Elliptic - Stabilises a vertex or vertices;
- Inversion - Reflection about the mid-point of an edge (Often also called elliptic);
- Hyperbolic - Translation along a bi-infinite path in the tree
My Question: Given a hyperbolic automorphism $\alpha \in \text{Aut}(\mathcal{T}_n)$ and a bi-infinite path $(v_k)_{k \in \mathbb{Z}} \subset V\mathcal{T}_n$, can a hyperbolic automorphism translating along $(v_k)_{k \in \mathbb{Z}}$ be obtained by sequentially composing $\alpha$ with a sequence $(\alpha_k)_{k=1}^{m} \subseteq \text{Aut}(\mathcal{T}_n)$ (i.e. $\beta = \alpha_1 \circ \alpha_2 \circ \cdots \circ \alpha_m \circ \alpha)$ of elliptic automorphisms of $\mathcal{T}_n$? And if so, can the new automorphism have the same translation distance of the original?
Fact: Every parity-preserving automorphism $\varphi$ of $\mathcal{T}_n$ (an automorphism that preserves the unique 2-coloring) can be written as a composition of two elliptic automorphisms. Consider any vertex $v$ of $\mathcal{T}_n$, and let $w$ be the midpoint between $v$ and $\varphi(v)$. Then there exists an automorphism $\psi$ fixing $w$ and sending $v$ to $\varphi(v)$, and $\psi^{-1} \circ \varphi$ fixes $v$. Hence $\varphi = \psi \circ (\psi^{-1} \circ \varphi)$ is the composition of two elliptic automorphisms.
Now let $\beta$ be any hyperbolic automorphism translating along $(v_k)$ with the same translation distance as $\alpha$. Since $\beta$ preserves parity iff $\alpha$ preserves parity, $\beta \circ \alpha^{-1}$ preserves parity, so it can be written as a composition $\alpha_1 \circ \alpha_2$ of two elliptic automorphisms. Now $\beta = \alpha_1 \circ \alpha_2 \circ \alpha$ as requested.
With a more elaborate argument you can show that $\alpha$ and $\beta$ are even conjugate.