Rational function on $\mathbb{P}^2$ whose square is identity

Question

Suppose $f:\mathbb{P}^2 \to \mathbb{P}^2$ is rational such that $f \circ f = \mathbb{Id}$. Then is it true that $f$ must be linear?

It feels true due to the degree which increases, but some things might cancel out.

Suppose we have a smooth curve $C$ of genus $g\geq 1$ with a rational function $g: C \to C$ with the same property. Does it always rise to an $f$ on $\mathbb{P}^2$ whose restriction is $g$? Does it imply that $g$ has to be linear too? This on the other side seems wrong to me.

Answer 1

As Mohan mentioned in the comments, there exist non-linear functions whose power is $\mathbb{Id}$.

For example $[x:y:z]\mapsto [yz:xz:xy]$ has order two, and $[x:y:z]\mapsto [xy:yz:zx]$ has order six.

Answer 2

Here is an involution

$$(x,y,z)\mapsto (x^2-yz, y^2-xz, z^2-xy)$$

It is the inversion wr to the conic $x y + x z + x z=0$ with inversion center $(1,1,1)$ ( the Hirst transform, see @Jean Marie: answer).

Projective rational involutions appear in other instances, think of the map $A \mapsto A^{-1}$ for matrices, which in projective coordinates can be given as $A \mapsto \operatorname{adj} A$, from a matrix to its adjugate. This map, restricted to certain subalgebras of matrices, again gives an involution.

Still, I cannot find any rational involutions of $\mathbb{P}^2$ of degree $>2$. Also, I can't find rational involutions of $\mathbb{P}^n$ of degree $> n$.