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I don't know a reference but maybe the following proof is shorter than yours. By continuity, $\varphi\circ\varphi_\infty=\varphi_\infty$. Hence $\varphi$ is identity on the set $I:=\varphi_\infty([0, …

Such simple adjustment is not possible. First, take the logarithm: consider new random variables $y_i=\log x_i$. Their ordering is the same, but correction is now additive rather than multiplicative: …

Here is another example which is easy to visualize. Let $X=[-1,1]^2$, $I$ be the segment between $(-1,0)$ and $(1,0)$ and $I_0$ its subsegment between $(-1/2,0)$ and $(1/2,0)$. There is a homeomorphis …

The answer is no, assuming that you seek an orientation preserving square root. (I see unknown's answer appeared while I'm typing. I don't quite understand it at the moment but the construction looks …

Here is an example. Consider a map $f:\mathbb R^2\to\mathbb R^2$ given by $(r,\varphi)\mapsto (r,\varphi+\sin(1/r))$ in polar coordinates $(r,\varphi)$, $0<r\le 1/\pi$. For $r>1/\pi$, let $f$ be ident …

Let me throw in some speculations based on my limited involvement in dynamical systems. The conjugation formula $f=h^{-1}gh$ is in general not a type of a functional equation that can be solved by it …

The general conjecture is false. Silly example: if $f(x)=2x$ and $g(x)=x^2$, then $f\circ f\circ g=g\circ f$. More interesting example: $f(x)=2x+1$ and $g(x)=x^2+2x$. It is obtained from the silly on …