The group of $k$-automorphisms of $k[x_1,\ldots,x_n,x_1^{-1}]$

Question

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". 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".

This question 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.

(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.

Answer 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.