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.