The following nonlinear integral transform takes continuous functions defined on the cylinder $\mathbb{R} \times S^1$ to $C^2$ functions defined on the plane $\mathbb{R}^2$:
$$ \mathcal{A}f (x,y) := \frac{1}{2\pi} \int_0^{2\pi} \! \! \! \int_0^\theta \sin(\theta - \phi)f(y\cos(\phi) - x\sin(\phi), \phi) f(y\cos(\theta) - x\sin(\theta), \theta) \, d\phi d\theta . $$
Question. Assume $f$ is strictly positive and is even in the sense that $f(-p, \theta + \pi) = f(p, \theta)$. Is it true that $\mathcal{A}f (x,y)$ is a constant if and only if $f(p, \theta)$ is independent of $p$? More generally, what can we say of two even positive functions that have the same transform?
Motivation. In the easiest possible case, Hilbert's fourth problem asks to construct and study the reversible Finsler metrics on the plane for which geodesics are straight lines. Hamel, a student of Hilbert, gave the following neat solution:
Theorem. (Hamel, 1903) All the geodesics of a reversible Finsler metric $F : T\mathbb{R}^2 \rightarrow [0, \infty)$ that is $C^2$ outside the zero section are straight lines if and only if there exists a continuous, strictly positive function $f : \mathbb{R} \times S^1 \rightarrow \mathbb{R}$ satisfying the symmetry condition $f(-p, \theta + \pi) = f(p, \theta)$ and for which $$ F(x,y,r\cos(\theta),r\sin(\theta)) = r\int_0^\theta \sin(\theta - \phi)f(y\cos(\phi) - x\sin(\phi), \phi) \, d\phi . $$
Recently, and among other things, I showed in this paper that in the case of the sphere, and in all dimensions, a reversible Finsler metric all of whose geodesics are great circles is completely determined by its volume form. The volume I'm referring to is the Holmes-Thompson volume (the pushforward onto the base manifold of the symplectic volume on the unit codisc bundle of the Finsler metric). Such a broad statement is not true in the plane: take two distinct norms such that the duals of their unit discs have the same (Euclidean) area, and you get two distinct Finsler metrics that have the same geodesics and the the Holmes-Thompson area form.
The relation between Holmes-Thompson area and the integral transform is made through the following simple
Proposition. The Holmes-Thompson area form of a a reversible Finsler metric $F : T\mathbb{R}^2 \rightarrow [0, \infty)$ that is given by Hamel's formula, $$ F(x,y,r\cos(\theta),r\sin(\theta)) = r\int_0^\theta \sin(\theta - \phi)f(y\cos(\phi) - x\sin(\phi), \phi) \, d\phi , $$ is $\mathcal{A}f (x,y) \, dx \wedge dy$.
In view of this, the OP asks whether norms are the only Finsler metrics on the plane for which both the unparametrized geodesics (i.e., Finsler solutions of Hilbert's fourth problem) and the area form agree with those of Euclidean geometry.
I have hunch that having the same unparametrized geodesics and the same volume form is extremely rigid even at the local level.
