Is there exist a similar conjecture to the famous [Jacobian Conjecture](https://en.wikipedia.org/wiki/Jacobian_conjecture) with $\mathbb{C}[x\_1,\ldots,x\_n,x\_1^{-1},\ldots,x\_n^{-1}]$ instead of 
$\mathbb{C}[x\_1,\ldots,x\_n]$?


Namely, let $f$ be $\mathbb{C}$-algebra endomorphism of $\mathbb{C}[x\_1,\ldots,x\_n,x\_1^{-1},\ldots,x\_n^{-1}]$, denote 
$f\_i:= f(x\_i)$, and further assume that the Jacobian of $\{f\_1,\ldots,f\_n\}$ is in $\mathbb{C}-\{0\}$. Is such $f$ an automorphism?


(I guess that first one should be familiar with the group of automorphisms of 
$\mathbb{C}[x\_1,\ldots,x\_n,x\_1^{-1},\ldots,x\_n^{-1}]$, see [this question](https://math.stackexchange.com/questions/2232869/the-group-of-k-automorphisms-of-kx-1-ldots-x-n-x-1-1-ldots-x-n-1)).


**Edit:** I also wonder if there exists any nice connection between the Jacobian Conjecture and my above conjecture (which is not exactly phrased yet); for example, are the two conjectures equivalent?



"Jacobian Conjecture" for $k[x_1,\ldots,x_n,x_1^{-1},\ldots,x_n^{-1}]$?
Counterexample: the endomorphism of the product of two punctured lines (complement of the curve of equation $xy=0$ in the plane) given by $$(x,y)\mapsto f(x,y)=\left(\frac{x}{y},y^2\right)$$


We have
$$\begin{pmatrix}\partial\_1f\_1 & \partial\_2f\_1\\ \partial\_1f\_2 & \partial\_2f\_2\end{pmatrix}=\begin{pmatrix}\frac{1}{y} & -\frac{x}{y^2}\\ 0 & 2y\end{pmatrix},$$
so the Jacobian is constant equal to 1.


It's not an automorphism, since both $(1,1)$ and $(-1,-1)$ map to $(1,1)$.


