Tree automorphisms without inversions are either elliptic or hyperbolic

By [L4], every displacement $d_T(v,gv)$ is a natural number, so choose $v$ with minimal displacement $m=d_T(v,gv)$. Then $m=\mathrm{\ell }(g)$ by [L2]. If $m=0$, the vertex $v$ is fixed and [L3] describes the fixed subtree.

Assume $m>0$, and let $P$ be the unique reduced path from $v$ to $gv$. Its translate $gP$ is the unique reduced path from $gv$ to $g^{2}v$. If $P$ and $gP$ met in more than the vertex $gv$, then some interior point of the overlap would have displacement strictly smaller than $m$, contradicting step 1.1; if they shared an edge with opposite orientations, that edge would be inverted, contradicting [L1]. Hence consecutive translates $g^{n}P$ and $g^{n+1}P$ meet only at one endpoint. [L1, L4, step 1.1, assume-case[hyperbolic], algebra]

Therefore $A_g:={\bigcup }_{n\in \mathbb{Z}}{g}^{n}P$ is a bi-infinite reduced path. It is preserved by $g$, and $g$ sends each segment $g^{n}P$ onto $g^{n+1}P$, so every vertex on $A_g$ moves distance exactly $m$ along that line. Thus $g$ acts on $A_g$ as translation by $\mathrm{\ell }(g)=m$.

Let $x$ be a vertex not on $A_g$, and let $y$ be the first vertex of $A_g$ on the unique reduced path from $x$ to $A_g$. Then the geodesic from $x$ to $gx$ runs from $x$ to $y$, then along $A_g$ from $y$ to $gy$ by length $m$, and then from $gy$ to $gx$, so $d_T(x,gx)=2d_T(x,A_g)+m>m$. Hence the vertices of minimal displacement are exactly those on $A_g$, which makes $A_g$ unique. This is the hyperbolic case, and it excludes fixed vertices.