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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.

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    You want to prove that there is a set of distinct vertices $v_i$ for $i \in {\mathbb Z}$ such that each $v_i$ is joined to $v_{i+1}$ by and edge (giving you a line), and such that there exists $k > 0$ with $gv_i = v_{i+k}$ for all $i$. You can take $k$ to be the minimum of $d(v,gv)$. – 2017-02-23
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    @DerekHolt So, I wanna show that there is a "infinite path" without end and beginning, which is invartiant under action by $g$ in a sense that $gv_i=v_{i+k}$? So should I try to look at the set $\{v:d(v,gv)=m\}$ where $m$ is the minimal $d(v,gv)$? – 2017-02-23
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    I would start from trying to understand how the proof could go from looking at a particular example ... – 2017-02-23
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    It's misleading to involve one group $G$ in the question that plays no further role... You have one isometry $g$ with some assumption and some conclusion, that's it. – 2017-03-02

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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)

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.