On the fixed point of automorphism of $\mathbb F_3[[T]]$ 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
 A: Let's write $R$ for $\mathbb{F}_3[[T]]$ and $K$ for its fraction field. Suppose $t \in R \setminus \mathbb{F}_3$ satisfies $\sigma(t)=t$. I claim that $\sigma$ then has finite order.
Suppose first that the valuation $k$ of $t$ is coprime to $3$. Then I claim that the order of $\sigma$ divides $k$. Indeed, there exists (after possibly taking a base extension of the ground field $\mathbb{F}_3$, as Michael Zieve pointed out in a comment) an automorphism $\alpha$ of $R$ that maps $t$ to $T^k$. So then $\alpha(\sigma(\alpha^{-1}(T)))=\zeta T$ for some $k$-th root of unity $\zeta$, hence $\sigma$ has order dividing $k$.
Now the general case. I am a bit uncertain about this, since it almost seems too easy, but here goes. I first want to prove that if $K_0$ is the topological closure of $\mathbb{F}_3(t)$ in $K$, then $K$ is of finite degree over $K_0$. Now $K_0$ is isomorphic to $\mathbb{F}_3((t))$, hence is a locally compact normed field with respect to the absolute value induced by that on $K$. Furthermore, the field $K=\mathbb{F}_3((T))$ is a locally compact normed $K_0$-vector space, so by e.g. Lemma 2 in these notes by Pete Clark) we conclude that $K$ has finite dimension over $K_0$. But then $\sigma$ extended to $K$ is an element of the finite group $\operatorname{Aut}(K/K_0)$. Therefore $\sigma$ has finite order as an automorphism of $K$, and therefore also as an automorphism of $R$.
Conversely, if $\sigma$ has finite order $k$, then symmetric expressions in $T$, $\sigma(T)$, $\ldots$, $\sigma^{(k-1)}(T)$ should give plenty of power series on which $\sigma$ acts trivially. (This has been worked out much better in Michael's answer and the comments below it.)
A: I show here: (1) if $c_1=1$ and $\sigma$ is not the identity then $\sigma$ has no fixed points whose lowest-degree nonconstant term has degree coprime to $3$; and (2) if $\sigma$ has finite order under composition, then $\sigma$ has fixed points outside $\mathbf{F}_3$.
To prove (1), let $t\in\mathbf{F}_3[[T]]$ be fixed by $\sigma$, and assume that the lowest degree of any nonconstant term of $t$ is some integer $n$ which is coprime to $3$. By definition, $\sigma(t)=t(T+f(T))$, where $f(T)\in\mathbf{F}_3[[T]]$ and the lowest degree of any term of $f(T)$ is some integer $m>1$.  By replacing $t$ by $t-t(0)$, we may assume that $t(0)=0$; this does not affect the other hypotheses on $t$.  Joe Silverman's argument handles the case $n=1$:  there is some $s\in\mathbf{F}_3[[T]]$ such that $s(t)=T$, so if
$t=\sigma(t)=t(T+f(T))$ then by applying $s$ to both sides we obtain the contradiction $T=T+f(T)$.  If $n>1$ then use Taylor expansion to compute
$$
\sigma(t)=t(T+f(T))=t+f(T)t'+\sum_{i=2}^{\infty} f(T)^i H^{(i)}(t),
$$
where $H^{(i)}(t)$ is the $i$-th Hasse derivative of $t$, defined by $H^{(i)}(\sum c_k T^k)=\sum c_k\binom{k}{i}T^{k-i}$.  But each term in the summation has degree at least $mi+n-i$, which is at least $m+n$ since $m\ge 2$.  Since the lowest-degree term of $f(T)t'$ has degree $m+n-1$, it follows that $\sigma(t)-t$ has a term of degree $m+n-1$ and hence $\sigma(t)\ne t$.
Item (2) was mostly shown in a (since-deleted) answer by user René, who observed that if $\sigma$ has finite order (say $n$), then it fixes every symmetric polynomial in $T, \sigma(T), \sigma^2(T), \dots, \sigma^{n-1}(T)$.  This proves (2) because the values of these symmetric polynomials can't all be in $\mathbf{F}_3$; this can be shown in various ways, for instance Yves Cornulier notes that the product of the $\sigma^i(T)$'s has a degree-$n$ term.
Finally, I note that there has been a good deal of work studying power series in $\mathcal{N}:=X+X^2\mathbf{F}_p[[X]]$ which have finite order under composition.  It isn't hard to show that any such element has order $p^r$ for some $r$.  Klopsch explicitly determined the elements of order $p$, up to conjugacy by an element of $\mathcal{N}$: they are $X(1-inX^n)^{-1/n}$ where $i\in\mathbf{F}_p^*$ and $n$ is a positive integer coprime to $p$.  The conjugacy classes of elements of order $p^r$ were described in terms of Artin-Schreier-Witt theory in Jean's thesis.  Alternate proofs and further developments are in Lubin's paper cited below.  Still, the problem of explicitly describing an element of order $p^r$ remains open in all cases except $r\le 1$ and $p=r=2$.  In case $p=r=2$ the element was constructed by Chinburg and Symonds; subsequently, together with Bleher and Poonen, they showed that their approach could not be generalized to other cases.  References are below.


* Benjamin Klopsch, Automorphisms of the Nottingham group, Journal of Algebra 223 (2000), 37-56

* Sandrine Jean, Conjugacy classes of series in positive characteristic and Witt vectors, Journal de Théorie des Nombres Bordeaux 21 (2009), 263-284

* Jonathan Lubin, Torsion in the Nottingham group, Bulletin of the London Mathematical Society 43 (2011), 547-560.

* Ted Chinburg and Peter Symonds, An element of order 4 in the Nottingham group at the prime 2, arXiv:1009.5135.

* Frauke Bleher, Ted Chinburg, Bjorn Poonen and Peter Symonds, Automorphisms of Katz-Gabber covers, preprint available online.

A: The elements in $\mathbb{F}_p[[T]]$ of the form $T+T^2f(T)$ form a group under composition, the Nottingham group $\mathcal{N}_p$. (See https://en.wikipedia.org/wiki/Nottingham_group.) So in the case that $c_1=1$, your $\sigma$ is in $\mathcal{N}_p$, and you're asking if there is some $\tau\in\mathcal{N}_p$ such that $\sigma\star\tau=\tau$. (Here I write $\star$ for the group law, which is composition.) Since $\mathcal{N}_p$ is a group, I can multiply (compose) on the right by $\tau^{-1}$ to conclude that $\sigma(T)=T$. Hmmm... Okay, so I guess this means that your $t$ needs to have a constant term, and also I guess you've assumed that your $f(T)$ is a polynomial, otherwise you can't evaluate $f(t)$ when $t$ is a power series. So this doesn't completely solve your problem, but it at least eliminates all $t$'s that don't have a constant term.
EDIT: As pointed out in the comments, what this argument eliminates is $t$'s of the form $T+T^2g(T)$. (And probably $cT+T^2g(T)$ with $c\ne0$). There remains the interesting question of $t$'s that start with a $T^2$ or higher terms. So my answer is merely a possible start, and also a suggestion that the literature on the Nottingham group might be relevant. The thesis of Matthew Gradner-Spencer might also be relevant:
https://repository.library.brown.edu/studio/item/bdr:11323/  He looks at various actions of the Nottingham group on power series starting $T^p+h.o.t.$.
A: 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 $S$.
