Consider $f(x)=mx$ and $g(x) =bx$ in which $m>0$ and $b>0$. Does there exist any homeomorphism $h$ such that $h \circ f$ = $g \circ h$ Consider $f(x)=mx$ and $g(x) =bx$ in which $m>0$ and $b>0$. Does there exist any homeomorphism $h$ such that $h \circ f$ = $g \circ h$? If exists find the domain of $h$.
 
Obviously I tried $h(mx)=bh(x)$. But I don't know how to go for solving it! In addition, I have no idea about how to prove homeomorphism existence in such problems.
 A: This community wiki solution is intended to clear the question from the unanswered queue.
The question does not provide information about domains and ranges of the functions $f, g$. So let assume $f, g : \mathbb R \to \mathbb R$. We ask  whether there exists a homeomorphism $h : \mathbb R \to \mathbb R$ such that $$(*) \quad h  \circ f = g \circ h .$$
Ned's comments show that a suggestive ansatz is
$$h(x) = \text{sgn}(x) \lvert x \rvert^L .$$
Here $L > 0$ is a real number which we can specify. Note that $h$ is in fact a homeomorphism since $L > 0$. (For $L \le 0$ the value $h(0)$ cannot not be defined by the above formula, but no matter which value we choose for $h(0)$: We never get a homeomorphism.)
$(*)$ is satisfied iff
$$m^L\text{sgn}(x)\lvert x \rvert^L = \text{sgn}(mx) \lvert mx \rvert^L = h(f(x)) = g(h(x)) = b\text{sgn}(x) \lvert x \rvert^L .$$
This is true iff $m^L = b$ which is equivalent to $L\ln m = \ln b$. Thus we must have


*

*either $\ln m = \ln b = 0$ which is equivalent to $m = b = 1$ in which case $L$ is arbitrary,

*or $\ln m, \ln b \ne 0$ which is equivalent to $m, b \ne 1$ in which case $L = \ln b / \ln m$. Note that to achieve $L > 0$ we must require that both $m, b > 1$ or both $m, b < 1$.
This shows that the ansatz works provided the above conditions on $m,b$ are satisfied.
What about the cases not covered by our ansatz? These are


*

*$m = 1, b \ne 1$

*$m \ne 1, b = 1$

*$m < 1, b > 1$

*$m > 1, b < 1$
Let us first observe that solutions of $(*)$ are never unique. In fact, if $h$ satisfies $(*)$, then also $h' = ch$ with $c \ne 0$ satisfies $(*)$ because $h'(f(x)) = ch(f(x)) = cg(h(x)) = cbh(x) = bch(x) =bh'(x) = g(h'(x))$. In case $m = b = 1$ we even see that each homeomorphism $h$ is a solution of $(*)$.
Let us verify that in cases 1. - 4. there does not exist a solution of $(*)$.


*

*If $m = 1$, then $f = id$ and we must have $h = h \circ f = g \circ h$, i.e. $g = id$ which is not true for $b \ne 1$.

*This is treated as 1.

*Assume $(*)$ has a solution $h$. Then $h(0) = h(f(0)) = g(h(0)) = b h(0)$ which implies $h(0) = 0$ since $b \ne 1$. We may assume that $h$ is strictly increasing (in case it is strictly decreasing, consider $h' = -h$ which is also a  solution). For $x > 0$ we have $0 < mx < x$, thus $0 = h(0) < h(mx) < h(x)$ which implies $h(f(x)) < h(x) < b h(x) = g(h(x))$, a contradiction.

*This is treated similarly as 3.
