Meaning of Jacobians satisfying $J_F^{-1}(J_F^{-1})^T = c\mathrm{Id}$ for a map $F$ between two domains? Let $F\colon \Omega_1 \to \Omega_2$ be a $C^1$ invertible map between smooth domains $\Omega_1, \Omega_2 \in \mathbb{R}^n$.
Suppose that the Jacobian matrix of $F$, called $J_F$, is invertible and satisfies the property
$$J_F^{-1}(J_F^{-1})^T = c\mathrm{Id}$$ for some constant function $c$. Of course the elements of $J_F$ and its inverse are functions of $x \in \mathbb{R}^n$.
Do such maps have some name attached to them and do they imply anything about the domains $\Omega_1$ and $\Omega_2$ (or $F$)?
 A: Your condition implies
$$c J_F^T = J_F^{-1} \Rightarrow (\sqrt c J_F)^T = (\sqrt c J_F)^{-1}$$
meaning $\sqrt c J_F$ is an orthogonal matrix. Notice that $\sqrt c J_F$ is the Jacobian of the function $G(x) = \sqrt c F(x)$.
A relevant mathoverflow post claims that the only functions $G$ with orthogonal jacobian are affine, i.e. of the form $G(x) = Ax+b$ for some constant orthogonal matrix $A$ and constant vector $b$. So,
$$F(x) = \frac{1}{\sqrt c} \left( Ax+b \right)$$
A: If $c>0$, such maps are called conformal if $\det J_F>0$ and anticonformal if $\det J_F<0$. Notice that I didn't specify a point at which $\det J_F$ is evaluated. That's because the sign can't change, which I will discuss later.
We will take a look at your function first, and then at conformal functions in general.
So you have a map $F$ with invertible Jacobian such that
$$J_F^{-1}(J_F^{-1})^T=c\operatorname{Id},$$
where $c\neq0$, since both factors on the left are invertible, so their product has to be invertible as well. Since $c>0$, we can rearrange this to
$$(\sqrt cJ_F)(\sqrt c J_F)^T=\operatorname{Id},$$
which makes $\sqrt c J_F$ orthogonal (see lisyarus' answer). In other words, $\sqrt c J_F$ represents a rotation, possibly combined with a reflection. Multiplying (=dilating) by $\sqrt c^{-1}$ gives us $J_F$, so $J_F$ is a rotation-dilation, possibly combined with a reflection. Another word for such a linear map is conformal (without reflection) or anticonformal (with reflection). They form the conformal group $CO(\mathbb R^n)$ (conformal orthogonal). A $C^1$ function with (anti)conformal Jacobian in $x_0$ is itself called (anti)conformal in $x_0$. It is called (anti)conformal if it is (anti)conformal on its entire domain. Intuitively, conformal and anticonformal functions are those which keep angles between smooth intersecting lines invariant. So a rectangular grid might be squished by such a function, but the intersections are still right angles. Wikipedia has a nice picture demonstrating this fact.
The conformal matrices are those in $CO(\mathbb R^n)$ with positive determinant, while those with negative determinant are anticonformal. This is what guarantees that your function $F$ is actually either conformal or anticonformal, not a mix between the two (like conformal at one point and anticonformal at another). If you consider the map $\varphi:\Omega_1\to\mathbb R\backslash\{0\},~x\mapsto\det J_F(x)$, then $\varphi(x)$ is either positive or negative, but never zero. Since $\varphi$ is continuous ($J_F(x)$ is continuous since $F$ is $C^1$ and the determinant is also continuous) and $\Omega_1$ is connected, the image of $\varphi$ is also connected. But then it must be either always positive or always negative, since the negative and positive part of $\mathbb R\backslash\{0\}$ are disconnected. So $J_F$ is conformal everywhere or anticonformal everywhere, and thus $F$ is either conformal or anticonformal.
An especially important example of conformal maps are holomorphic functions (with derivative nowhere zero): A function between domains in the complex plane is conformal in $z_0$ iff it is complex differentiable with non-zero derivative in $z_0$. That's essentially what the Cauchy-Riemann equations tell us: they guarantee that the Jacobian is conformal or zero.
