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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'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 $\mathbf{F}_3$, since if that happened then all the coefficients of $$ (X-T)(X-\sigma(T)(X-\sigma^2(T))\dots (X-\sigma^{n-1}(T)) $$ would be in $\mathbf{F}_3$, contradicting the fact that $T$ isn't a root of any nonzero polynomial in $\mathbf{F}_3[X]$.

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.

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.

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), note that $\sigma(t)=t(T+f(T))$ where 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 whether $\sigma$ fixes $t$. Let $n$ be the lowest degree of any term of $t$, so $3\nmid n$. Joe'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 $\mathbf{F}_3$, since if that happened then all the coefficients of $$ (X-T)(X-\sigma(T)(X-\sigma^2(T))\dots (X-\sigma^{n-1}(T)) $$ would be in $\mathbf{F}_3$, contradicting the fact that $T$ isn't a root of any nonzero polynomial in $\mathbf{F}_3[X]$.

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.

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'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 $\mathbf{F}_3$, since if that happened then all the coefficients of $$ (X-T)(X-\sigma(T)(X-\sigma^2(T))\dots (X-\sigma^{n-1}(T)) $$ would be in $\mathbf{F}_3$, contradicting the fact that $T$ isn't a root of any nonzero polynomial in $\mathbf{F}_3[X]$.

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.

I show here that if $\sigma$ has finite order under composition, then $\sigma$ has fixed points outside $\mathbf{F}_3$.

This was mostly shown in a (since-deleted) answer by user René, who noted that if $\sigma$ has finite order (say $n$), then it fixes every symmetric polynomial in $T, \sigma(T), \sigma^2(T), \dots, \sigma^{n-1}(T)$. Note that the values of these symmetric polynomials can't all be in $\mathbf{F}_3$, since if that happened then all the coefficients of $$ (X-T)(X-\sigma(T)(X-\sigma^2(T))\dots (X-\sigma^{n-1}(T)) $$ would be in $\mathbf{F}_3$, contradicting the fact that $T$ isn't a root of any nonzero polynomial in $\mathbf{F}_3[X]$.

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.

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), note that $\sigma(t)=t(T+f(T))$ where 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 whether $\sigma$ fixes $t$. Let $n$ be the lowest degree of any term of $t$, so $3\nmid n$. Joe'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 $\mathbf{F}_3$, since if that happened then all the coefficients of $$ (X-T)(X-\sigma(T)(X-\sigma^2(T))\dots (X-\sigma^{n-1}(T)) $$ would be in $\mathbf{F}_3$, contradicting the fact that $T$ isn't a root of any nonzero polynomial in $\mathbf{F}_3[X]$.

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.