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?

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?

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?

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?