I try to find the automorphisms $\sigma$ of $\mathbb F\_q((\frac1T))$ with the following properties: $\sigma$ is an isometry,and $\sigma(\mathbb F\_q[T])\subseteq\mathbb F\_q[T]$. Almong the automorphisms
$$\mathcal N=\{\sigma(\frac1T)=\sum\_{n\ge1} \alpha\_i\frac1{T^i}\mid\alpha\_i\in\mathbb F\_q \}$$ the Nottingham group of $\mathbb F\_q((\frac1T))$, I only found the trivial automorphisms $T\mapsto \alpha T$. Does there exist other ones (maybe not in $\mathcal N$)?



Automorphisms of $\mathbb F_q((\frac1T))$
There are a few more. Suppose $\sigma(T) =a\_n T^n + \dots + a\_0$. Then


$$ \sigma \left(\frac{1}{T}\right)= \frac{1}{\sigma(T)} = a\_n^{-1} T^{-n} \left(1- \frac{a\_{n-1}}{a\_n} T^{-1} + \frac{a\_{n-1}^2 - a\_n a\_{n-2}}{a\_n^2}T^{-2} + \dots \right)$$


This is not an automorphism unless $n=1$. In that case the formula looks like:


$$ \sigma\left(\frac{1}{T} \right) = \frac{1}{a\_1} \frac{1}{T} - \frac{a\_0}{a\_1^2} \frac{1}{T^2} + \frac{a\_0^2}{a\_1^3} \frac{1}{T^3} + \dots $$


or more elegantly:


$$ \sigma\left(\frac{1}{T} \right) = \alpha \frac{1}{T} + \alpha \beta \frac{1}{T^2} + \alpha \beta^2 \frac{1}{T^3} + \dots $$


Theser are all.


