Action of homeomorphism on real line An element $f ∈ Homeo^+(\mathbb{R})$ is said to be of :
(1) type A, if it has a trivial germ at ∞ and it does not fix any point in an
interval of the form (−∞, r).
(2) type B, if it has a trivial germ at −∞ and it does not fix any point in an
interval of the form (s, ∞).
(3) type C, if it does not fix any point in intervals of the form (−∞, r) and
(s, ∞) for some r < s, and fixes the points r, s.
(4) fully supported if it does not fix any points in $\mathbb{R}$
Question:
Let $f$ be an element of type A, B or C, and let $g ∈ Homeo^+(\mathbb{R})$
be a fully supported element. Is it possible for some $n ∈ \mathbb{Z}$, the homeomorphisms
$g^{−n}fg^n$ and $f$ commute?
 A: This is not possible. I will write a proof for type A; the proofs for types B and C are similar. Without loss of generality, $n \in \mathbb Z_{>0}$.
By the intermediate value theorem, we must either have $g(x) > x$ for all $x$, or $g(x) < x$ for all $x$. Replacing $g$ by $g^{-1}$ if necessary (and changing $f$ accordingly), we may assume $g(x) < x$ for all $x$. Similarly, replacing $f$ by $f^{-1}$ if necessary (we don't need to change $g$), we may assume $f(x) < x$ for all $x \ll 0$.
Assume $f$ does not fix any point in $(-\infty,r)$ and is the identity on $(s,\infty)$. Then the closed set
$$V = \{x \in \mathbb R\ |\ f(x) = x\}$$
is nonempty and bounded below, hence has a minimal element $x_0$. Moreover, again by the intermediate value theorem, we have
$$f(x) < x \hspace{1cm} \text{ for all } x < x_0\label{1}\tag{1}$$
(since this holds for $x \ll 0$ and $x_0$ is the smallest fixed point). If $h = g^{-n}fg^n$ commutes with $f$, then we get
$$fh(x_0) = hf(x_0) = h(x_0),$$
hence $y_0 = h(x_0)$ is fixed by $f$. But $g^n(x_0) < x_0$, hence $fg^n(x_0) < g^n(x_0)$ by (\ref{1}), hence
$$y_0 = g^{-n}fg^n(x_0) < x_0$$
since $g^{-n}$ is strictly increasing. We conclude from (\ref{1}) that $f$ does not fix $y_0$. $\square$
