The group of $k$-automorphisms of $k[x_1,\ldots,x_{n},x_1^{-1},\ldots,x_n^{-1}]$ Let $k$ be a field (I do not mind to further assume that $k$ is of zero characteristic, if that will make things easier).
For $k[x_1,\ldots,x_n]$ it is known that the affine and triangular automorphisms generate the group of automorphisms of $k[x_1,\ldots,x_n]$, call it $G_n$,
see, for example, van den Essen's book.
It is also known that $G_2$ is a free amalgamated group, see, for example, Dick's paper.
My question:
Is the group of $k$-automorphisms of $k[x,x^{-1}]$, call it $\hat{G_1}$, known?
of $k[x,y,x^{-1},y^{-1}]$?
or more generally, of $k[x_1,\ldots,x_n,x_1^{-1},\ldots,x_n^{-1}]$?
(call it $\hat{G_n}$).
By 'known' I mean if it is not difficult to find a set of generators similarly to the generators of $k[x_1,\ldots,x_n]$ (or perhaps, a free amalgamated structure of $\hat{G_2}$ similarly to the free amalgamated structure of $G_2$).
Thanks for any hints and comments.
 A: These automorphism groups are much easier to understand than the automorphism groups of polynomial rings.  Take $A=k[x,x^{-1}]$.  If $\varphi:A\to A$ is an endomorphism, then $\varphi(x)$ is a unit of $A$.  By degree considerations, it is easy to see that the only units of $A$ are monomials $ax^n$ for $a\in k^\times$ and $n\in\mathbb{Z}$.  For $\varphi$ to be surjective, we must have $n=\pm 1$.  Conversely, for any $a\in k^\times$ and $n=\pm 1$, $x\mapsto ax^n$ does give an automorphism.  Explicitly, then, the group is a semidirect product of $k^\times$ ($x\mapsto ax$) and $\mathbb{Z}/2$ ($x\mapsto x^{-1}$), with $\mathbb{Z}/2$ acting on $k^\times$ by inversion.
With more variables the story is similar but more complicated.  An endomorphism of $k[x_1,\dots,x_n,x_1^{-1},\dots,x_n^{-1}]$ must send each $x_i$ to a unit, and the only units are $ax_1^{d_1}\dots x_n^{d_n}$ for $a\in k^\times$, $d_1,\dots,d_n\in\mathbb{Z}$.  Such an endomorphism is an automorphism iff the exponents $(d_1,\dots,d_n)$ for each of the $x_i$ form a basis of the abelian group $\mathbb{Z}^n$.  It follows that the automorphism group is a semidirect product of $(k^\times)^n$ (automorphisms of the form $x_i\mapsto a_i x_i$) and $GL_n(\mathbb{Z}$) (automorphisms of the form $x_i\mapsto x_1^{d_{i1}}\dots x_n^{d_{in}}$ where $(d_{ij})\in GL_n(\mathbb{Z})$).
