A group acting on tree without inversion acts on a line as a translation? I'm asked to show:

If $G$ acts on a tree $T$ without inversions (so $ge\neq \bar{e}$ for every edge $e$ and $g\in G$, where $\bar{e}$ is edge in reverse direction), and $g\in G$ fixes no vertex in $T$, then $g$ acts on some line $L\subset T$ by translation.

But unfortunately I don't quite understand question, nor do I have any idea how I would start. First of all, what is a "line"? I think I'm supposed to guess what it means (does it, probably, mean a set of edges that are "connected"?). When it says "acts as translation" I guess it means that it takes a path $v_1v_2\dots v_n$ to $(gv_1)\dots(gv_n)$?
The hint says that start by considering vertices $v$ such that $d(v,gv)$ is minimal, but I'm not sure where I should start from.
 A: I think I have a solution now.
Let $k=\min_{v\in V(T)} d(v,gv)$. Because $g$ fixes no vertex, $k$ must be a positive integer. Now set $L=\{v\in V(T):d(v,gv)=k\}$. Pick a vertex $v_0\in L$, and let $v_0\to v_1\to\cdots \to v_{k-1}\to v_k=gv_0$ be a path of minimal length $k$.
Because $T$ is a tree, and as this path must be reduced, this path is a unique path of length $k$ joining $v_0$ to $gv_0$.
Now consider $v_1$. Via group action, the edge $(v_0,v_1)$ is mapped to $(gv_0,gv_1)$. We claim that $gv_1\neq v_i$ for $i=0,\dots,k$
$gv_1\neq v_0$ because if it were, we have formed a circuit, but $T$ is a tree. A special note here is that if $k=1$ then there wouldn't be a circuit, but then in that case $g$ acts on the edge $(v_0,v_1)$ via inversion, which is forbidden.
$gv_1\neq v_1$ since $g$ cannot fix a vertex.
For $i\ge 2$ we note that if $gv_1=v_i$ then $d(v_1,gv_1)<k$ via already estabilished path, contradicting minimality of $k$.
Therefore, $v_1$ is also in $L$ with the path of minimal length $v_1\to v_2\to \cdots\to v_k\to gv_1$.
Thus inductively, if $L$ contains a vertex then it must contain infinitely many vertices forming a line, contained in a single infinite path, and by above $g$ acts on $L$ by translation.
