Is normalizing automorphism in $\mathbb{R}^n$ by Jacobian still an automorphism?

Question

Let $n \in \mathbb{N}$. Suppose $f: \mathbb{R}^n \to \mathbb{R}^n$ is a $C^\infty$ diffeomorphism. Let $J_f: \mathbb{R}^n \to \text{M} (n, \mathbb{R})$ be the Jacobian of $f$. Is the map $F: \mathbb{R}^n \to \mathbb{R}^n$ defined by $F(x) = [J_f(x)]^{-1}\cdot f(x)$ a $C^\infty$ diffeomorphism?

Answer 1

I don't think this is even true for $\mathbb{R}$.

Consider the function $x^3 + e^x$. This function is clearly smooth and bijective, and the inverse is smooth by repeated application of the inverse function theorem, as the derivative has no zeros, and the inverse function theorem can be iterated to arbitrarily many derivatives.

However, for us here, the Jacobian is just the derivative, and your map $F$ is just $f/f'$. But for us, this is $\frac{x^3 + e^x}{3x^2 + e^x}$ which is not injective. In particular, it attains the value $1$ at $0$, but also at $x = 3$