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, $\prod_{\gamma\in\Gamma}\gamma(T)$. So if $u$ is a torsion element of Nottingham, then the group $\langle u\rangle$ is just such a $\Gamma$ as above.

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 $S$.