Let $k$ be a field (of characteristic zero).
For $k[x\_1,\dotsc,x\_n]$ it is known that the affine and triangular automorphisms generate $G\_n$, the group of automorphisms of $k[x\_1,\dotsc,x\_n]$, 
see, for example, [van den Essen's book "Polynomial automorphisms and the Jacobian conjecture"](http://www.springer.com/gp/book/9783764363505).
It is also known that $G\_2$ is a free amalgamated group, see, for example, [Dicks's paper "Automorphisms of the polynomial ring in two variables"](http://www.raco.cat/index.php/PublicacionsSeccioMatematiques/article/viewFile/37473/37347).


[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) asks what is $\hat G\_n$, the group of automorphisms of $k[x\_1,\dotsc,x\_n,x\_1^{-1},\dotsc,x\_n^{-1}]$.


Now, my question is:



> 
> What is $\tilde G\_n$, the group of automorphisms of $k[x\_1,\dotsc,x\_n,x\_1^{-1}]$? Namely, we only invert one variable, say $x\_1$.
> 
> 
> 


Thus far, what I was able to obtain is as follows:


(1) $k[x,x^{-1}]$: This case was already done in the answer to the above mentioned [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).


(2) $k[x,x^{-1},y]$: If $f$ is an automorphism, then it is necessary that $f(x)$ is invertible in $k[x,x^{-1},y]$, so (if I am not wrong), $f(x)=\lambda x^d$, for some non-zero scalar $\lambda$ and some integer $d$. In order to be surjective, $d \in \{\pm1\}$. Then, unless I am missing something, $f(y)$ must be of the following form: $\varphi(y)= \mu y + \sum\_{i=s}^{t}c\_i x^{i}$, where $\mu$ is a non-zero scalar, $c\_i$ are scalars, and $s \leq t$ are integers. So there are only ‘a few’ automorphisms,
which are similar to the usual triangulars (the difference is the existence of negative exponents for $x$).


(3) Perhaps the $n \neq 3$ is similar to the $n=2$ case.



The group of $k$-automorphisms of $k[x_1,\ldots,x_n,x_1^{-1}]$
I repeat my comment as an answer as suggested.


Your argument for $n=2$ generalises and shows that $x\_1$ is always mapped to some $\lambda x\_1^m$ for $\lambda\in k^\times$ and $m\in\{\pm1\}$. Therefore if you consider the subring $S:=k[x\_1^{\pm1}]$, a general $k$-linear automorphism of $R:=k[x\_1^{\pm 1},x\_2,\ldots,x\_n]$ is a composition of a $S$-linear automorphism of $S[x\_2,...,x\_n]$ and a $k$-linear automorphism of $S$ (acting on $S[x\_2,...,x\_n$] by acting on coefficients). Phrased slightly differently
$$Aut\_k(R) = Aut\_S(S[x\_2,\ldots,x\_n])\rtimes Aut\_k(S) = Aut\_S(S[x\_2,\ldots,x\_n]) \rtimes (k^\times \rtimes \{\pm1\})$$


But note that $Aut\_S(S[x\_2,\ldots,x\_n])$ is not the group $G\_{n-1}$, only a subgroup of it (and for a different field of coefficients: $k(x\_1)$ instead of $k$), because $S$ is not a field. Nevertheless, this gives you a lot more information than you had before.


