$\mathbf {The \ Problem \ is}:$ It is given that the set of all possible homeomorphisms on $\mathbb R$ forms a group $(\mathbb H_{\mathbb R},.)$ under functional composition. Then prove that the two functions $f_1(x)=2x$ and $f_2(x)=3x$ lie in the same conjugacy class in $(\mathbb H_{\mathbb R},.).$
$\mathbf {My \ approach} :$ Actually, by thinking backwardly if there exists some $g \in \mathbb H_{\mathbb R}$ such that $f_1=gf_2g^{-1}$, then letting $g^{-1}=h$, we have $h(2x)=3h(x)$ for all $x \in \mathbb R$, which means $\frac {h(2x)-h(x)}{2}=h(x)$ where $h$ is either strictly increasing or strictly decreasing on $\mathbb R$ and $h(0)=0.$ I think we have to use ivp of $h$, but I can't approach properly . And, obviously $h$ is not a polynomial in $x.$
A small hint is warmly appreciated .
Hint: Try conjugating with the map $$g(x) = \cases{ x^t, & if $x\geq 0$ \\ -|x|^t, & if $x \leq 0$}$$ for suitable $t$.