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Let $M$ be a smooth manifold, and $f,g:M\rightarrow M$ be topologically conjugate(i.e., there exists a homeomorphism $h:M\rightarrow M$ such that $f\circ h=h\circ g$). If $f,g$ are differentiable, is it possible that $f,g$ have different order of differentiability?

TCK17
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    $x \mapsto x^3$ and $x \mapsto x \cdot |x|$ on $\mathbb{R}$ are topologically conjugated. – D. Thomine Aug 22 '16 at 14:45
  • @ D. Thomine, can you show me the conjugate transformation? – TCK17 Aug 22 '16 at 14:55
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    There is a standard recipe to construct them, see e.g. here. Work separately on $(-\infty, -1]$, $[-1,0]$, $[0,1]$ and $[1, +\infty)$. Check that each conjugation fixes the endpoints (because $-1$ and $1$ are repelling for both $f$ and $g$, and $0$ is attracting for both), so you can glue the four conjugations you get. – D. Thomine Aug 22 '16 at 15:06

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