Can we wrap a square onto itself with constant singular values? I have now cross-posted this on mathoverflow.
Let $0<\sigma_1<\sigma_2$ satisfy $\sigma_1\sigma_2=1$, and let $D=[-1,1]^2$.

Does there exist a Lipschitz bijective* map $f:D \to D$ such that $df$ has almost everywhere the fixed singular values $\sigma_1,\sigma_2$?

Is there such a diffeomorphism of $D$? (thinking of $D$ as a manifold with corners, or requiring smoothness only on the interior etc.)
*I am fine with requiring only $|f^{-1}(y)|=1$ for a.e. $y \in D$; the Area formula then implies that $f$ is surjective.
Clearly, no affine map would be suitable. We somehow need a map whose singular vectors are 'rotating' from point to point.

Comment: If we replace $D$ with a disk then we have $ f_t:(r,\theta) \to (r,\theta+t \log r)$, which
is the flow of $\log r \frac{\partial}{\partial \theta}$. $f_t$ has constant singular values (which depend on $t$.)
 A: Here are some elements of reflection, not at all a full answer.
Let us think into the complex plane, using conformal mappings (i.e. bi-holomorphic bijections). [seeing your $\theta+c\log(r)$, we are not far anyway from  complex numbers...]
There exists explicit conformal mappings from the square $D$ (any square in fact) onto unit disk $U$. Let for example $\varphi$ be the one given here using complex function  $$\varphi(u):=\int_0^u \dfrac{dz}{(z^4-1)^{1/2}}\tag{1}$$
(itself issued from Schwarz-Christoffel transformation).
Caution: in this case the vertices are $\{1,i,-1,-i\}$ instead of $\pm1\pm i$.
There exists an infinity of conformal mappings: $\mu:U \to U$, under the form of these Möbius transformations:
$$\mu_{\theta,a}(z)=\mu(z)=e^{i\theta }{\frac{z+a}{\bar{a}z+1}}, \ \ |a|<1. \tag{2}$$
Then, you can take $$\varphi^{-1}\circ \mu \circ \varphi: D \to D \tag{3}$$
Turning (3) into its equivalent in terms of real functions ($x+iy \to (x,y)$), it may give you, using chain rule, a way to attack your issue.
But I am fully conscious that this does not bring a direct answer to your query...
