Does the data of Galois group, ramified places, and inertia groups, determine a Galois number field? Suppose I tell you that $K/\mathbb{Q}$ is a finite Galois extension, and I specify the Galois group $G$, and suppose further that I give you a finite list $S$ of places of $\mathbb{Q}$ and for each place $p$ a conjugacy class $C_p$ of subgroups of $G$, and I tell you that $K/\mathbb{Q}$ is ramified precisely at the places $p$ of $S$, with inertia groups $C_p$.
Is this enough information to recover $K$?
If not, can you provide a counterexample? If so, can you refer me to some sources (or, if the proof is not too hard, can you indicate lines of proof)?
ADDENDUM August 8: To clarify a point. The list $S$ is allowed to include the infinite place. If the infinite place is ramified, i.e. if $K$ is not a real field, then, fixing some embedding of $K$ in $\mathbb{C}$, $G$ contains complex conjugation and the relevant conjugacy class $C_\infty$ is that of the subgroup generated by complex conjugation. So, the data $S,\{C_p\}$ that we start with includes whether the infinite place is ramified.
 A: I would say no. For example, take $K=\mathbb Q(i)$ and $L=\mathbb Q(\sqrt{2})$, which are non-isomorphic number field. Both are ramified precisely at $2$, and the inertia subgroup at $2$ is the whole Galois group. So if I tell you $G=C_2$, $S=\{2\}$ and $C_2=G$, you are not able to recover exactly one between $K$ and $L$.
A: Ferra's $K$ and $L$ are distinguished by the infinite place, but they inspired me to look for simple quadratic counterexamples:
Take $K=\mathbb{Q}(\sqrt{3})$ and $L=\mathbb{Q}(\sqrt{6})$. These are both ramified at 2 and 3 and because the Galois group of both is $\mathbb{Z}/2\mathbb{Z}$, whose only nontrivial subgroup is the whole group, the inertia groups at both 2 and 3 must be the whole group.  Then $K$ and $L$ cannot be distinguished by the information specified in the question.
Actually, for any squarefree $m$ that is $>0$ and $3$ mod $4$, $K=\mathbb{Q}(\sqrt{m})$ and $L=\mathbb{Q}(\sqrt{2m})$ would also always provide a counterexample, because the discriminants are $4m$ and $8m$ and these are divisible by the same primes, so they are ramified at the same places.
