The group of $k$-automorphisms of $k[[x,y]]$, $k$ is a field

Let $k$ be a field. Is the group of $k$-automorphisms of $k[[x,y]]$ known? ($k[[x,y]]$ is the ring of formal power series in two variables, see Wikipedia.)

A somewhat relevant question is this question, which deals with $k[[x]]$, with $k$ any commutative ring.

Thanks for any hints and comments.

It is simple to describe all automorphisms of $R=k[[x,y]]$ as a $k$-algebra. First, note that $R$ is a local ring with maximal ideal $m=(x,y)$, so any automorphism must map $m$ to itself and thus be continuous in the $m$-adic topology. Since $k[x,y]$ is dense in $R$ in the $m$-adic topology, this means that an automorphism is determined by where it sends $x$ and $y$. Moreover, given $f,g\in m$, there is a unique continuous $k$-algebra homomorphism $\varphi:R\to R$ such that $\varphi(x)=f$ and $\varphi(y)=g$ (given by "evaluating" power series by plugging in $f$ for $x$ and $g$ for $y$). So the only question is what conditions on $f$ and $g$ guarantee that this $\varphi$ is an automorphism.

The answer is simple: you just need the images of $f$ and $g$ in the $k$-vector space $m/m^2$ to be linearly independent. Concretely, this just means that the linear homogeneous parts of $f$ and $g$ are linearly independent. Clearly this condition is necessary, since the images of $x$ and $y$ in $m/m^2$ are linearly independent. Conversely, if the linear homogeneous parts of $f$ and $g$ are linearly independent, it is easy to show $\varphi$ is surjective (for any homogeneous polynomial $p$ of degree $n$, you can find a homogeneous polynomial $q$ of degree $n$ such that $p$ is the degree $n$ part of $\varphi(q)$, and then you can use this to build a power series whose image under $\varphi$ has any desired value, one degree at a time). A surjective homomorphism from a Noetherian ring to itself is automatically injective, so it follows that $\varphi$ is an automorphism.

So automorphisms of $R$ are in bijection with pairs of power series with no constant term whose linear parts are linearly independent. Beware that the group operation is very complicated from this description--to compose two automorphisms, you need to compose the power series (that is, substitute the power series of one automorphism for the variables $x$ and $y$ in the power series of the other automorphism).