Domination of length functions of trees with equal covolume (This is a reformulation of an earlier unanswered question. I would like to thank Ian Agol for pointing out to me Walter Parry's characterization of hyperbolic translation length functions.)
Let $G$ be a finitely generated group and $T_1$ and $T_2$ two minimal (and hence cocompact) metric simplicial trees that $G$ acts on isometrically by simplicial automorphisms without inversions. Consider the (unbased) hyperbolic translation length functions $$l_{T_i}\colon G\to [0,\infty),\ g\mapsto\inf_{x\in T_i}d(x,gx),\ i\in\left\{1,2\right\}.$$
Assume that $T_1$ and $T_2$ have the same covolume, say covolume 1. Moreover, suppose that for all $g\in G$ we have $l_{T_1}(g)\leq l_{T_2}(g)$ and $l_{T_1}(g)=0\Leftrightarrow l_{T_2}(g)=0$.
Does this imply that $l_{T_1}$ and $l_{T_2}$ are equal?
I have the feeling that for every $g$ such that $l_{T_1}(g)$ is strictly smaller than $l_{T_2}(g)$ there must be an $h\in G$ such that $l_{T_1}(h)$ is strictly bigger than $l_{T_2}(h)$ in order to preserve the covolume. Indeed, I can prove the statement for free $F_n$-trees (elements of Culler-Vogtmann Outer space), but my arguments do not generalize immediately.
 A: Suppose $S$ is a metric $G$-tree of covolume $1$ (ie the sum of the edge lengths in $G\backslash S$ is equal to 1).  Now suppose that $T$ is obtained from folding together two adjacent edges in the same $G$-orbit, as follows:
Let $e$ be an edge of $S$ adjacent to a vertex $v$, and suppose that $g\in G$ stabilizes $v$ but not $e$.  Then $T$ is obtained from $S$ by ($G$-equivariantly) folding $e$ and $ge$.
This operation doesn't change the underlying graph of the quotient, so $T$ still has covolume equal to 1.
Consider a hyperbolic element $\gamma\in G$.  The axis of $\gamma$ in $T$ is contained in the image of the axis of $\gamma$ in $S$, so $l_T(\gamma)\leq l_S(\gamma)$.  However, if the axis of $\gamma$ in $S$ traversed a pair of edges in $S$ that are folded in $T$, then we will have a strict inequality 
$l_T(\gamma)< l_S(\gamma)$ .
For a specific example, let $G$ be the free group $\langle a,b\rangle$, and let $S$ be the Bass--Serre tree of the HNN extension
$G\cong \langle a\rangle*_1$
where $b$ is the stable letter.  Now let $T$ be the Bass--Serre tree of the HNN extension
$G\cong \langle a,a^b\rangle*_{a\sim a^b}$
where, again, $b$ is the stable letter.  Note that the commutator $[a,b]$ has strictly smaller translation length in $T$ than in $S$---indeed, it's hyperbolic in $S$ but elliptic in $T$. 
A: Here is a counterexample (unfortunately):
We give the two graphs of groups the same metrics. Equivariantly folding the two branching edges in the Bass-Serre covering $T_1$ gives rise to an equivariant 1-Lipschitz map from $T_1$ to $T_2$ and hence we have $l_{T_2}(g)\leq l_{T_1}(g)$ for all $g\in G$. Now observe that the graph of groups of $T_2$ can be obtained from the graph of groups of $T_1$ by an elementary collapse followed by an elementary expansion. Elementary deformations do not create new elliptic subgroups, whence $l_{T_1}(g)=0\Leftrightarrow l_{T_2}(g)=0$.
However, if we give each edge length $1/3$, the hyperbolic group element corresponding to the upper loop followed by a backtrack along the connecting bridge has translation length $1$ in $T_1$ and length $1/3$ in $T_2$.
The two trees are in fact non-abelian (they are irreducible), as the fundamental groups of their graphs of groups contain a free group of rank 2 acting freely.

A: I guess this is proved in F.Paulin, The Gromov topology on $\mathbb R$-trees, Topology Appl. 32 (1989).
Abstract:

We are interested in isometric actions of a fixed finitely generated
  group on R-trees. Using metric methods inspired by Gromov’s work, we
  define a more geometric topology on sets of such objects. We prove it
  to be the same as the Morgan-Shalen topology, defined by the
  translation lengths of the group elements, in the case of minimal
  irreducible actions.

