Let $k[x_1,\ldots,x_n]$ be a polynomial ring over a field $k$ of characteristic zero.

When $n=2$, it is known that every automorphism of $k[x_1,x_2]$ is tame, namely, a finite product of elementary automorphisms.

For $n=3$, in their paper, Shestakov and Umirbaev showed that the Nagata map is wild (=non-tame, not belongs to the subgroup generated by elementary automorphisms).

My question: For $n \geq 3$, if we know that a given automorphism $g$ is of finite order (namely $g^m=1$ for some $m$), must it be tame?

Or is it possible to have a wild automorphism of finite order?

Sorry if my question is trivial.

Polynomial Automorphisms and the Jacobian Conjectureit is written that it is not known whether finite order automorphisms are linearizable (an affirmative answer to which will imply an affirmative answer to the Cancellation problem: does $V \times \mathbb{C} \cong \mathbb{C}^{n+1}$ imply $V \cong \mathbb{C}^n$?). It seems nonetheless that most believe the answer to be negative: there should be wild polynomial involutions. $\endgroup$ – Vesselin Dimitrov Jan 17 '16 at 21:25