action by isometries on tree is constant on conjugacy classes A real $F_n$-tree is a real tree $T$ (a metric space with distance $d_T$ such that each two points can be connected by a unique geodesic) on which $F_n$ (the free group of rank $n$) acts by isometries (i.e. for $g \in F_n$ and $x,y \in T$ we have $d_T(gx, gy) = d_T(x,y)$). These are definitions from "The Tits alternative for $\operatorname{Out}(F_n)$ II: A Kolchin type theorem" by Bestvina, Feighn and Handel.
They also define a length function: suppose the distance on the tree is given by $d_T$. Let $a \in F_n$ and define the length
$$l_t(a) = \operatorname{inf}_{x \in T}\{d_T(x, ax)\}$$
where $ax$ is the image of $x$ under the action of $F_n$. The text claims that this lenght is constant on conjugacy classes. I tried proving this, but failed: my idea was to show that $d_T(x, ax) = d_T(x, gag^{-1}x)$ for each $x \in T$ and $a, g \in F_n$. I tried using the fact that $F_n$ acts by isometries, but did not manage to 'cancel terms'. 
Next the text also claims that $\{x \in T \vert d_T(x, ax) = l_T(a)\}$ is a set isometric to $\mathbb{R}$. I can not seem to find the required 'map'. Any hints would be appreciated!
 A: The first property is not special to $\mathbb{R}$-trees or free groups.  Suppose that $G$ is a group acting on the left by isometries on a complete metric space $(X,d)$.  The translation length of an element $g \in G$ is defined as 
$\ell_X(g) = {\rm inf} \, \{d(x, g.x) \mid x \in X\}$.
If $h \in G$, then $d(x,h^{-1}gh.x) = d(h.x,gh.x)$ because $h$ acts by isometries.
Taking the infimum over all $x \in X$ is the same as taking the infimum over all $y \in X$ such that $y = h.x$ for some $x \in X$ because for any $y$, we have $y = h.(h^{-1}.y)$.
Therefore, $\ell_X(h^{-1}gh) = \ell(g)$.
The second statement does use some geometry and properties of the group.  For $g \in G$, the set $\{x \in X \mid d(x, g.x) = \ell_X(g)\}$ is called the min set of $g$, denoted by ${\rm Min}(g)$ in the text by Bridson and Haefliger, a good reference and a good place to learn more.
If $x \in {\rm Min}(g)$ and $\ell_X(g) > 0$, then $g^n.x \in {\rm Min}(g)$ for all $n \in \mathbb{Z}$ because $d(g^n.x, g.(g^n.x)) = d(x, g.x) = \ell_X(g)$.  Notice if $gh = hg$, then $h.x \in {\rm Min}(g)$ by the same argument.  But in a free group, you won't have this additional complexity.
Let's assume from now on that $X$ is an $\mathbb{R}$-tree and $G$ is a group acting freely by isometries.  Let $[x,g.x]$ denote the unique unit speed geodesic from $x$ to $g.x$.  If $y \in [x,g.x]$, then $y \in {\rm Min}(g)$ because 
$$d(x,y) + d(y,g.x) = d(x,g.x) = \ell_X(g) = d(g.x, g.y) + d(y,g.x) = d(y,g.y)$$.
It follows that the union $L$ of the geodesics $[g^n.x, g^{n+1}.x]$ for $n \in \mathbb{R}$ belongs to the min set.  On the other hand, if $z$ does not belong to $L$, then there is a unique closest point $\pi(z)$ in $L$.  Since $X$ is an $\mathbb{R}$ tree, $d(z, g.z) = 2*d(z, \pi(z)) + \ell_X(g)$.  Therefore, $L = {\rm Min}(g)$.  And $L$ is isometric to $\mathbb{R}$.  I've omitted some details.
The final important detail is why ${\rm Min}(g)$ is non-empty and $\ell_X(g) > 0$ if $g \neq 1$.  The exercise, see (3) on p.231 of Bridson & Haefliger, is that if $x \in X$ and $m$ is the midpoint of $[x,g.x]$, then $m$ belongs to the min set.  Since the group is acting freely, $g.x \neq x$ and so $\ell_X(g) > 0$.
