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Consider the group $\text{Aut}\mathbb{R}$ of smooth invertible maps from $\mathbb{R}$ to $\mathbb{R}$. If $f\in\text{Aut}\mathbb{R}$ has order 2 ($f$ is an involution), is $f$ conjugate to $g(x)=-x$?

The answer is "yes" if we replace the word "smooth" with "continuous" or "continuously differentiable", but I do not know how to resolve this in the case of smooth automorphisms.

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    $\begingroup$ This is widely known as (smooth) diffeomorphism group. It's true. One way to prove this is to observe that $g$ preserves a Riemannian metric, and then use that every Riemannian manifold that is homeomorphic to $\mathbf{R}$ is smoothly isometric to an interval of $\mathbf{R}$. By the existence of an orientation-reversing involution, it has to be smoothly isometric to $\mathbf{R}$ or $]-1,1[$. One easily concludes. $\endgroup$
    – YCor
    Commented Oct 15, 2019 at 13:07

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