Group action on the real line I was wondering about the following question: 
if you have a faithful action of a group $G$ on the real line $\mathbb{R}$ by orientation-preserving homeomorphisms, it is easy to construct a new action such that a point $p$ in $\mathbb{R}$ has the trivial stabilizer in $G$. Is it possible to make (possibly by a completely new action) the map $G \to \mathbb{R}$ defined by $g \mapsto g(p)$ is a group homomorphism always (when $\mathbb{R}$ is regarded as a group with addition)? If not, when could it be done? 
Any comment and/or advice would be greatly appreciated. 
Edit: As many people pointed out already, if such a homomorphism $G \to \mathbb{R}$ exists, then $G$ must be abelian — I was over-simplifying the question. What I really want is 

when G is a countable group acting faithfully on $\mathbb{R}$ and $S$ is a finite subset of $G$, can I construct a new action so that a point $p$ with trivial stabilizer satisfies that $g(p) + h(p) = g(h(p))$ for all $g, h\in S$?  

 A: Just to expand on Will's answer, prompted by Yves' comment to it.
If $G$ has an orientation-preserving action on $\mathbb{R}$ with trivial stabilisers, then the action is rigid in the sense that no compact interval $[x,y]$ is mapped properly into itself by any $g$ --- otherwise $g^n x$ would converge to a point fixed by $g$. It follows that each $g\neq 1$ is either positive ($gx>x$ for all $x$) or negative ($gx<x$ for all $x$). This defines an order on $G$ which is both left- and right-invariant. (Rigidity is necessary to guarantee right-invariance.) 
I claim that this order is archimedean, meaning that for $g,h\in G$ with $h\neq 1$ there exists $n\in\mathbb{Z}$ with $g<h^n$. Otherwise, $x\leq h^n x<gx$ for all $n\geq 0$ which would imply a fixed point for $h$.
So $G$ is an archimedean ordered group. Thus by an old result of H\"older, $G$ is abelian, and in fact can be naturally embedded in $\mathbb{R}$ along the lines of Will Sawin's answer. And it is not hard to show that (additive) subgroups of $\mathbb{R}$ are indeed either discrete or dense.
EDIT (following comments to the original question)
To cover all bases, suppose that $G$ acts on $\mathbb{R}$ and that
the stabiliser of $p$ is trivial, but the same is not necessarily
true of all points. Then focusing on the action of $G$ on the orbit
$Gp$, we still obtain a two-sided order on $G$. Now if
$\phi:g\mapsto g\cdot p$ is to be a homomorphism (for a possibly new
action) then since the target is abelian
the commutator subgroup $[G,G]$ must be contained in the
kernel of the (new) action. So for a faithful action, $G$ must be
abelian. But not all (even finitely generated) orderable groups are
abelian; for example the soluble Baumslag-Solitar group $BS(1,n)$
acts on the line with (at least some) trivial stabilisers: the maps
$x\mapsto x+1$ and $x\mapsto nx$ generate a group isomorphic to
$BS(1,n)$. But these groups are of course not abelian for $n\geq
2$.
A: This is true (edit: I mean here: the existence of an action with one point with trivial stabilizer) when $G$ is countable, or more generally when it acts faithfully order-preserving on a countable totally ordered set:  
1) If an arbitrary group $G$ acts order-preserving faithfully on a totally ordered set $(D,\le)$, then there exists a left-invariant total ordering $\le'$ on $G$ (so that the action of $G$ on itself by left-translation is free and order-preserving)
Indeed, consider a well-ordering $\preceq$ on $D$ (unrelated to the total order), and define $g\le' h$ if $g=h$ or $g\neq h$ and the $\preceq$-minimal element $y$ of $\{x\in D:g(x)\neq h(x)\}$ satisfies $g(y)\le h(y)$.
2) If an arbitrary group acts order-preserving on a countable totally ordered set $(D,\le)$, then this action can be extended to an action on the reals.
Indeed, first let $G$ act on $D\times ]0,1[$ with the lexicographic ordering (so that each $\{d\}\times ]0,1[$ is convex) by $g(d,t)=(gd,t)$, and extend the action to the Dedekind cut completion. The latter is isomorphic as a totally order set to the real line.
3) Using 1) and 2), if a countable group acts faithfully order-preserving on a totally ordered set, then it also admits a faithful action on the real line with a free orbit (i.e., so that at least one point has a trivial stabilizer).
On the other hand, I'm not sure that the group $\text{Homeo}^+(\mathbf{R})$ admits an order-preserving action on $\mathbf{R}$ with a free orbit.
(edit: obviously, as pointed out by other people, if $G$ is not abelian it's hopeless to expect to realize this action by a homomorphisms into the reals)
A: It can be done when every point has a trivial stabilizer. Divide into two cases - $\{g(p) | g \in G\}$ dense and discrete.
In the first case $G$ is a totally ordered group whose order completion is homeomorphic to $\mathbb R$, since, as a totally ordered set, it is isomorphic to  $\{g(p) | g \in G\}$. It's ordered completion must clearly be the group $\mathbb R$, the unique complete Archimedean totally ordered group. But this exactly defines an order-preserving isomorphism, thus a homeomorphism between $\mathbb R$ the group and $\mathbb R$ the target of the action of $G$.  Thus the action of $G$ on $\mathbb R$ is homoemorphic to its action on $\mathbb R$ the group, which is a homomorphism.
In the second case $G = \mathbb Z$. Fix any homoemorphism $\mathbb R/G = \mathbb R/\mathbb Z$, and lift to an isomorphism of the original $\mathbb R$ to an $\mathbb R$ where the action of $G$ is a homomorphism.
If one point has a nontrivial stabilizer, this provides an upper or lower bond on $g(p)$, so it can never be the image of a homomorphism. 
