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 $g, h\in S$?

someoreverypoint $p$ has a trivial stabilizer. This dramatically changes the answer! $\endgroup$