Decomposition of primes in Galois closures of number fields Let $L/K$ be an extension of number fields, and $M/K$ the Galois closure of $L/K$ (everything happens inside a suitably large characteristic zero field $\Omega$). Let $p$ be a discrete prime of $K$.
Given the group $G={\rm Gal}(M/K)$, its subgroup $H={\rm Gal}(M/L)$, and the splitting of $p$ in $L/K$ can one find the splitting of $p$ in $M/K$? That is, can one find the ramification index $e$ and inertial degree $f$ of a prime $P$ of $M$ lying above $p$?

Turning the above question into a group-theoretic one, I got the following: let $D$ be a finite group, and $X$ a finite set on which $D$ acts faithfully on the right. Can we obtain the order $d$ of $D$ knowing the sizes of all the $D$-orbits in $X$?
(To switch to this question from the original one, look at the natural right action of a decomposition group (resp. an inertia group) $D$ at $P$ on the coset space $X=H\backslash G$. The fact that $M$ is the Galois closure of $L$ ensures that $D$ acts on $X$ faithfully.)
 A: Changed answer: The answer is no: Let $L$ be a root field of $X^3-2$ over $\mathbb Q$, and $M$ the Galois closure of $L/\mathbb Q$. Then the primes $2$ and $3$ are both totally ramified in $L$, yet in both cases there is only one prime $P$ above $p=2$ or $p=3$, with $e(P)=3$ if $p=2$, and $e(P)=6$ if $p=3$.
A: With regard to the group-theoretic question, the answer is no: for example, there are plenty of faithful transitive group actions of different groups on the same set (i.e. only one orbit, but it doesn't determine the size of the group). For instance, take $S_n$ and $A_n$ acting on $\{1,2,\dots, n\}$, as soon as $n \geq 3$.
A: Peter's answer made me want to modify the original question by adding the residue characteristic of $p$ to the list of numerical data in the question's hypotheses.  Here is an alternative example where the residue field of $p$ is fixed.
Let $f(x)=x^4-4x^2+2$ (so that $f(x+1/x)=x^4+1/x^4$).  Then, if $c$ is $4$, $8$, or $10$, adjoining a root of $f(x)-c$ to $\mathbb{Q}$ yields a degree-$4$ extension $L/\mathbb{Q}$ in which the rational prime $p=2$ is totally ramified.  Moreover, the Galois closure $M/\mathbb{Q}$ of $L/\mathbb{Q}$ is a degree-$8$ extension whose Galois group is dihedral.  Let $q$ be the unique prime of $L$ lying over $2$.  If $c=4$ then $q$ is inert in $M/L$; if $c=8$ then $q$ splits completely in $M/L$; and if $c=10$ then $q$ is totally ramified in $M/L$.
A: The answer is no. Let $K=\mathbb{Q}$ (it does not really matter) and let $M/\mathbb{Q}$ be a degree $6$ extension with Galois group $S_3$; finally, pick for $L$ any of the three cubic subfields of $M$.
By Chebotarev, there are infinitely many primes in $M$ splitting completely in $M/\mathbb{Q}$ and there are infinitely many unramified primes in $M$ whose decomposition subgroup is $\mathrm{Gal}(M/L)$. Then you cannot distinguish them if you only live in $L$ but their splitting behaviors in $M$ are different.
