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Dec 16, 2013 at 4:27 comment added Joe Silverman @MichaelZieve: I just edited my answer to include a link to Matt's thesis. He was mostly concerned with the conjugation action, but did prove a few things about the left and right actions. I'm not sure how much is relevant.
Dec 16, 2013 at 4:25 history edited Joe Silverman CC BY-SA 3.0
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Dec 16, 2013 at 4:00 comment added Joe Silverman @Julian and Mike: Good point. I guess all that I eliminated was $t$'s of the form $T+h.o.t$. But maybe that's a start. My student, Matt Gardener-Spencer, looked at the action of the Nottingham group on the set of power series that look like $T^p+h.o.t.$. And someone else (I forget who at the moment) has done further work on that. Which might or might not be relevant. In any case, I thought the OP should know that papers on the Nottingham group might be helpful.
Dec 16, 2013 at 1:36 comment added Michael Zieve this only shows that no fixed points have a nonzero coefficient of $T$, not that there are no fixed points outside $\mathbb{F}_3$.
Dec 16, 2013 at 1:33 comment added Michael Zieve Joe, constant terms don't affect the question, since they're fixed and $\sigma$ is a continuous homomorphism. Keep in mind that $\sigma$ is defined by $\sigma(\tau):=\tau(u)$ for any $\tau\in\mathbb{F}_3[[T]]$, where $u:=c_1 T + f(T)$. Crucially, $\sigma$ evaluates an arbitrary power series at a power series with no constant term, not the other way around. So $\tau$ is fixed if and only if $\tau(u)=\tau$, and your argument says that if $\tau'(0)\ne 0$ then we obtain $u=T$ by subtracting $\tau(0)$ from both sides and then composing on the left with the inverse of $\tau-\tau(0)$. But (cont.)
Dec 16, 2013 at 0:32 comment added Julian Rosen I think $t$ could fail to have a constant term if the linear term of $t$ is also $0$.
Dec 16, 2013 at 0:28 history answered Joe Silverman CC BY-SA 3.0