Is there an area-preserving concentric diffeomorphism of the ellipse? $\DeclareMathOperator\Vol{Vol}$This is a cross-post.
Let $0<b<1$ be a fixed parameter, and let $(R(\theta),\theta)$ be the polar coordinates of the ellipse
$$E=\{(x,y) \in \mathbb R^2 \, | \,  \Bigl(\frac{x}{1/b}\Bigr)^2+\Bigr(\frac{y}{b}\Bigr)^2=1\}.$$
Set
$$D=\{(x,y) \in \mathbb R^2 \, | \,  0<\Bigl(\frac{x}{1/b}\Bigr)^2+\Bigl(\frac{y}{b}\Bigr)^2 \le 1\}=\bigcup_{0<r\le1}r E,$$
and let $f:D \to D$ be given in polar coordinates by
\begin{equation}
\label{1}
\bigl(\tilde r  R(\theta),\theta\bigr )\mapsto \biggl(\psi(\tilde r)R\bigl(\theta+h(\tilde r)\bigr),\theta+h(\tilde r)\biggr). \tag{1}
\end{equation}
$f(\tilde rE)=\psi(\tilde r)E$; the "scale" is given by $\psi$ and the "phase" is given by $h$. If I am not mistaken, the Jacobian of $f$ is given by
$$
Jf=\frac{\psi(\tilde r)\psi'(\tilde r)}{R(\theta)}R\bigl(\theta+h(\tilde r)\bigr)\biggl(\frac{R\bigl(\theta+h(\tilde r)\bigr)}{\tilde r R(\theta)}+\frac{R'(\theta)}{R^2(\theta)}h'(\tilde r) \biggr).
$$
Question: Do there exist a $C^1$ monotone bijection $\psi:(0,1] \to (0,1]$, and a $C^1$ function $h:(0,1] \to \mathbb{R}$ that satisfy $Jf=1$ everywhere, besides $\psi(x)=x$ and $h=0$ and $h=\pi$ (which correspond to $f=\operatorname{Id}$ and to $f(x,y)=-(x,y)$).
If $h$ is constant, then it easily follows that $h=0$ or $h=\pi$ and $\psi(x)=x$, so I am looking for solutions with non-constant $h$.

Comment: $$R(\theta)\mathrel{:=}\frac{b}{\sqrt{1-\bigl(e \cos(\theta)\bigr)^2}}, e=\sqrt{1-b^4}.$$
 A: This is a tentatve (since I am not sure I am parsing your query correctly) comment, but too long for that format.  I take it that you are asking about the existence of diffeomorphisms of the interior of an ellipse which are area-preserving  and preserve the foliation consisting of dilations of the boundary.  The answer to this question is--yes, there are a multitude.  The rather imprecise formulation is that if you specify two distinct radial vectors on the one hand and two suitable curves from the centre to the boundary, then there is a unique such diffeomorphism which maps the former pair of curves to the latter.
There are two practical problems which are (from a mathematical point if view) identical to yours (stripped of the specifics)

*

*thermodynamics--identifying possible sysems of isotherms and adiabats of a thermodynamical system;


*cartography--identifying possible systems of parallels and meridians of area-equivalent map projection.
In both cases, the rough and ready answer is as follows: you can choose one of the systems at random (think hyperbolae $xy=c$ for isotherms ($T=pV$!) and $y=c$ for parallels).  One can then choose two adiabats, resp., two meridians at will and this determines the whole configuration.
There doesn't seem to be much literature on this but for the thermodynamics you could try the arXiv papers 1102.1540, 1108.4758.
Edit:  If we use new variables $u$ and $v$ where $u=(b^2 x^2+y^2/b^2)/2$ and $v$ is the angle $CAB$ where $A$ is the origin, $B$ is $(1,0)$ and $C$ is $(bx,y/b)$ (these are just polar coordinates in the circle case $b=1$ adjusted to make  the transformation area-preserving ), then the required transformations have the form $$(u,v) \mapsto (u,v+f(u))$$ for suitable functions $f$. (This is basically the solution of Robert Bryant but without using heavy machinery).
A: Here is another way to look at it that you might find useful:  Essentially, you are asking the following problem:  Given a cylinder $C = (0,R)\times S^1$ and a positive $2$-form $\omega = f(r,\theta)\,\mathrm{d}r\wedge\mathrm{d}\theta$, one wants to describe the diffeomorphisms of $C$ that preserve the $2$-form $\omega$ and the foliation given by the level curves of $r$.  For simplicity, I will assume that one only cares about the symplectomorphisms of $(C,\omega)$ that fix the two 'ends' and that the total $\omega$-area of $C$ is finite; I'll say what to do about the larger group and the infinite area case at the end.
First, make a change of variables:  Let
$$
\rho = \frac1{2\pi}\int_0^r\int_0^{2\pi} f(\tau,\theta)\ \mathrm{d}\tau\wedge\mathrm{d}\theta
$$
Then $\rho:(0,R)\to (0,P)$ is a diffeomorphism, where $2\pi P\le\infty$ is the total $\omega$-area of $C$, and we can regard $\rho$ as a function on $C$ with the same level sets as $r$.  Obviously, any symplectomorphism of $(C,\omega)$ that preserves the given foliation (and the ends of $C$) will preserve $\rho$, so one might as well replace $r$ by $\rho$ (and set $R=P$).  Thus, let's just assume that $\rho=r$.
One now has that
$$
\int_0^r\int_0^{2\pi} f(\tau,\theta)\ \mathrm{d}\tau\wedge\mathrm{d}\theta
= \int_0^r\int_0^{2\pi} \mathrm{d}\tau\wedge\mathrm{d}\theta,
$$
and, differentiating with respect to $r$, this gives
$$
\int_0^{2\pi} \bigl(f(r,\theta)-1\bigr)\ \mathrm{d}\theta  = 0
$$
for all $r$.  In particular, it follows that, setting
$$
\phi(r,\theta) = \int_0^\theta \bigl(f(r,\psi)-1\bigr)\,\mathrm{d}\psi,
$$
one has $\phi(r,\theta+2\pi) = \phi(r,\theta)$, so that $\phi$ is well-defined on $C$.  Moreover,
$$
\mathrm{d}r\wedge\mathrm{d}\phi = \bigl(f(r,\theta)-1\bigr)\,\mathrm{d}r\wedge\mathrm{d}\theta,
$$
so $\mathrm{d}r\wedge\mathrm{d}(\theta+\phi) = f(r,\theta)\,\mathrm{d}r\wedge\mathrm{d}\theta = \omega$.  Setting $\psi = 
\theta+\phi$, one now has $\omega = \mathrm{d}r\wedge\mathrm{d}\psi$.
In the coordinates $(r,\psi)$ on $C$, it is now easy to describe the symplectomorphism of $(C,\omega)$ that preserve the foliation defined by $\mathrm{d}r=0$, they are just the diffeomorphisms of the form
$$
F(r,\psi) = \bigl(r,\psi+g(r)\bigr),
$$
where $g$ is an arbitrary (differentiable) function of $r$.  Strictly speaking, these are the ones that preserve the ends of $C$.  To get the ones that exchange the ends of $C$ while preserving the given foliation, one can compose an $F$ of the above form with the involution
$$
H(r,\psi) = (R-r,-\psi).
$$
Now, in the case that the $\omega$-area of $C$ is infinite, one has to slightly modify the construction by integrating from a fixed $r=r_0\in(0,R)$ instead of from an 'end'.  I.e., one defines
$$
\rho = \frac1{2\pi}\int_{r_0}^r\int_0^{2\pi} f(\tau,\theta)\ \mathrm{d}\tau\wedge\mathrm{d}\theta.
$$
Then $\rho:(0,R)\to\mathbb{R}$ is a diffeomorphism onto its image, which might be all of $\mathbb{R}$.  In fact, if the image is not all of $\mathbb{R}$, then at least one of the two 'ends' has finite area and one can reduce to the case that the image is either $(0,\infty)$ or $(-\infty,0)$ and the arguments go through essentially unchanged.  When the image is all of $\mathbb{R}$, one can still replace $r$ by $\rho$ as a coordinate on $C$ and the one still has
$$
\int_0^{2\pi} \bigl(f(r,\theta)-1\bigr)\ \mathrm{d}\theta  = 0,
$$
and again, everything goes through as before, but now, when one gets to the coordinate system $(r,\psi)$ in which $\omega = \mathrm{d}r\wedge\mathrm{d}\psi$, the group of diffeomorphisms that preserve the level sets of $r$ and the $2$-form $\omega$ are now generated by
$$
F(r,\psi) = \bigl(r+c,\psi+g(r)\bigr)
$$
where $c$ is any constant and $g$ is any (smooth) function of $r$ and
$$
H(r,\psi) = (-r,-\psi).
$$
