I have done this same exercise for $Z_4$-set which I've found it way more intuitive. I know I have to think about this in an "abstract" way but I'm pretty lost. I do have the answer to it but still don't understand it.
Useful knowledge: 1) Any transitive $S_3$-set is isomorphic to the $S_3$-set of left cosets of some subgroup of $S_3$. 2) Two $S_3$-sets of left cosets of subgroups H and K are isomorphic if and only if they are conjugate.
Thoughts: Since $S_3$ has only 6 subgroups, there are at most 6 transitive $S_3$-sets. 3 of the subgroups of $S_3$ have order two (and can prove they are conjugate) which by 2) can conclude that there are exactly 4 transitive $S_3$-sets up to isomorphism.
Now the real issue appears when I want to make the tables (for example the group applying to a set of 6 elements a,b,c,d,e,f):
How was this table constructed?
As I said, doing these kind of tables is intuitive with $Z_4$ but not with $S_3$.
Thanks a lot