TITLE: On the fixed point of automorphism of $\mathbb F_3[[T]]$
QUESTION [8 upvotes]: Consider the automorphism $\sigma$ on ${\Bbb F}_3[[T]]$ such that $T \mapsto c_1T + f(T)$ with $c_1 = 1$ or $-1$, and $f(T) \not=0$ and the non-zero leading term $c_mT^m$ of $f(T)$ satisfies $m \geq 2$. 
Question: Is there any fixed element $t \in {\Bbb F}_3[[T]]$ other than those in the constant field ${\Bbb F}_3$? Namely does such $t$ exist as $\sigma(t) = t$ 
but $t \notin {\Bbb F}_3$?
Pierre

REPLY [2 votes]: Let’s call $K=k((T))$, where $k$ is a field of characteristic $p>0$. Suppose $\Gamma$ is any finite subgroup of the group of $k$-automorphisms of $K$, any one such necessarily sending $T$ to $u(T)=\sum_1^\infty a_iT^i$, and thereby sending a general element of $K$, say $g(T)=\sum_?^\infty b_iT^i$, to $g\circ u=\sum_?^\infty b_iu^i$. Then if $|\Gamma|=n$, of course its fixed field $E\subset K$ has $[K\colon E]=n$. The extension will always be totally ramified, and with a nice generator equal to the norm of $T$, that is, $\prod_{\gamma\in\Gamma}\gamma(T)$. If we call this series $S$, then $E=k((S))$. Now if $u$ is a torsion element of Nottingham, then the group $\langle u\rangle$ is just such a $\Gamma$ as above, and we get a whole lot of fixed elements under right composition by $u$, all of them power series (or Laurent series, if you wish) in $S$.