$\newcommand{Vol}{\text{Vol}}$
This$\DeclareMathOperator\Vol{Vol}$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 $$E=\{(x,y) \in \mathbb R^2 \, | \, (\frac{x}{\frac{1}{b}})^2+(\frac{y}{b})^2=1\}. \, \, \, \text{ Set}$$ $$D=\{(x,y) \in \mathbb R^2 \, | \, 0<(\frac{x}{\frac{1}{b}})^2+(\frac{y}{b})^2 \le 1\}=\cup_{0<r\le1}r E,$$ and $$E=\{(x,y) \in \mathbb R^2 \, | \, \Bigl(\frac{x}{1/b}\Bigr)^2+\Bigr(\frac{y}{b}\Bigr)^2=1\}.$$ 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
$$ \big(\tilde r R(\theta),\theta\big )\mapsto \bigg(\psi(\tilde r)R\big(\theta+h(\tilde r)\big),\theta+h(\tilde r)\bigg). \tag{1}$$ \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 $$ Jf=\frac{\psi(\tilde r)\psi'(\tilde r)}{R(\theta)}R\big(\theta+h(\tilde r)\big)\bigg(\frac{R\big(\theta+h(\tilde r)\big)}{\tilde r R(\theta)}+\frac{R'(\theta)}{R^2(\theta)}h'(\tilde r) \bigg). $$$$ Jf=\frac{\psi(\tilde r)\psi'(\tilde r)}{R(\theta)}R\bigl(\theta+h(\tilde r)\bigr)\biggl(\frac{R\bigl(\theta+h(\tilde r)\bigr)}{\tilde r R(\theta)}+\frac{R'(\theta)}{R^2(\theta)}h'(\tilde r) \biggr). $$
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,\pi$$h=0$ and $h=\pi$ (which correspond to $f=\text{Id}$$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: $$R(\theta):=\frac{b}{\sqrt{1-\big(e \cos(\theta)\big)^2}}, e=\sqrt{1-b^4}.$$$$R(\theta)\mathrel{:=}\frac{b}{\sqrt{1-\bigl(e \cos(\theta)\bigr)^2}}, e=\sqrt{1-b^4}.$$