This is a follow-up of this question.
Let $U \subseteq \mathbb R^2$ be an open, bounded, connected subset.
Let $f:U \to U$ be a smooth map. Suppose that $Jf=1$ ($f$ is area-preserving) and that $h:=f \circ f$ is affine (i.e. $\nabla^2h=\operatorname{Hess}h=0$).
Is $f$ affine?
If we omit the requirement that $Jf=1$, then there are counter-examples:
Take in the complex numbers the annulus $U = \{ z ∈ ℂ;~\frac 1 2 < \lvert z \rvert < 2\}$ and $f \colon U → U,~z ↦ \frac 1 z$.
However, no holomorphic counter-examples are possible, if we insist on area-preservation:
Since holomorphic maps are conformal, and conformality+Jacobian $1$ implies isometry (up to a constant scaling) every holomorphic area-preserving maps is affine.
As @fedja suggests in the comments, we construct an example as follows in polar coordinates:
$$ f:(r,\theta) \mapsto (\sqrt{1-r^2},-\theta). $$
This maps $U \to U$ when $U=\{ z ∈ ℂ\, | \, 0.6 < \lvert z \rvert < 0.8\}$.
$f$ is an involution, i.e. $f^2=Id$ is affine, but $f$ is not affine