= Solution
For a finite point $p$, use the <valuation> $v_p$ with local parameter $u=t-p$; at infinity use $u=1/t$. Its extensions $w$ to $E$ are permuted transitively by $G$. The completed residue <field> is $F$, because $F$ is algebraically closed, so the local <decomposition group> equals its <inertia group>.
Write $e=e(w/v_p)$. <Characteristic zero> makes all local ramification tame. One can choose a <uniformizer> $\pi$ in the completion for which
$$
u=\pi^e,\qquad E_w=F((\pi)),\qquad F(t)_{v_p}=F((u)).
$$
Indeed, initially $u$ is $\pi^e$ times a unit; that unit has an $e$th root in $F[[\pi]]$, by its algebraically closed residue <field> and the invertibility of $e$. Absorb it into $\pi$. The local automorphisms are then $\pi\mapsto\zeta\pi$, $\zeta\in\mu_e(F)$, and form a cyclic group of order $e$. Define $e_p=e$.
To specify an element rather than a subgroup, \b[fix a compatible choice of primitive <roots of unity> $\zeta_e\in F$]. Define $\tau_w$ by its <uniformizer> character
$$
\overline{\tau_w(\pi)/\pi}=\zeta_e,
$$
and let $C_p$ be its <conjugacy class> in $G$. Over $\mathbb C$, take $\zeta_e=e^{2\pi i/e}$, corresponding to positive local continuation around $p$. At an unramified point, $e_p=1$ and $C_p=\{1\}$.
Changing the <uniformizer> does not affect this character. If $\pi'=a\pi$ with $a$ a unit, inertia fixes its residue, so
$$
\overline{\tau(\pi')/\pi'}
=\overline{\tau(a)/a}\,\overline{\tau(\pi)/\pi}
=\overline{\tau(\pi)/\pi}.
$$
The character is injective: a nonidentity finite-order automorphism with derivative one would have $\tau(\pi)=\pi+b\pi^r+\cdots$ for $r\geq2$, and its $N$th iterate would have coefficient $Nb$ at that first nonzero displacement. <Characteristic zero> contradicts finite order. Thus the chosen primitive character identifies a unique generator.
For another place $w'=\sigma w$, conjugation gives $D_{w'}=\sigma D_w\sigma^{-1}$, preserves its order, and carries $\tau_w$ to $\tau_{w'}$, since $\sigma$ fixes the constants and hence $\zeta_e$. This proves that $e_p$ and the <branch-cycle conjugacy class> are independent of the place and <uniformizer>. Without the stated roots-of-unity convention, only the cyclic inertia-subgroup <conjugacy class> is intrinsic: replacing $\zeta_e$ by another primitive root replaces the generator by a coprime power, which need not lie in the same <conjugacy class>.
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