If I know the Conjugacy classes of a group, do I know the group? I know that a group has Conjugacy classes of size 1, 3, 6, 6, 8 and I know that this matches with the Conjugacy classes of the group $S_4$. But could there be a different group, with the same Congucy classes?
 A: I think that in general this question is difficult. Sometimes the answer is negative (see Zev Chonoles' answer). In your specific example case we can conclude that the group has to be $S_4$.
The elements in the conjugacy class of $8$ elements have centralizers of order $24/8=3$. Therefore they must be of order $3$, and generators of their own centralizers. Thus we can deduce that the group, call it $G$, has four Sylow $3$-subgroups. The group acts on the set of these four groups. This gives us a homomorphism $f:G\to S_4$.
The conjugation action of $G$ on the four Sylow $3$-subgroups is transitive by Sylow's theorem. We can further deduce that the action is doubly transitive. This is because each Sylow 3-subgroup of $G$ permutes the other three. Thus the group $f(G)$ is a doubly transitive subgroup of $S_4$. The only doubly transitive subgroups of $S_4$ are $S_4$ and $A_4$. In the former case we are done, because $f$ is then an isomorphism. So the claim follows, if we can prove that it is impossible for 
$\operatorname{Im} f$ to be isomorphic to $A_4$. Assume contrariwise that this would be the case. Then we could deduce that $\ker f$ is of order two. Because $\ker f$ is a normal subgroup of $G$ this implies that both elements of $\ker f$ would be in a conjugacy class by itself. This was manifestly not the case.
A: I'm afraid I don't know the answer to your specific question about $S_4$, but in general, knowing the number and sizes of the conjugacy classes is not sufficient, because any two finite abelian groups of the same cardinality $n$ will both have $n$ conjugacy classes of size $1$, but they need not be isomorphic. For an explicit case, consider $\mathbb{Z}/4\mathbb{Z}$ and $(\mathbb{Z}/2\mathbb{Z})^2$.
