I think I have the correct answer but I just want to double check. I think one of the solutions is {e,$r^2$,$r^4$,j,$r^2$j,$r^4$j}. I believe the second one is {e, $r^2$,$r^4$,r j,$r^3$j,$r^5$j} but I'm a bit unsure if my solutions are correct. Any insight would be helpful.
You are right. But, it is easy to describe geometrically. Imagine a regular hexagon. Its has six rotations and six reflections (symmetries). They form dihedral group of order 12.
In hexagon, you can see two "regularly sited figures". Consider three alternate edges (there are two possibilities; which help us to obtain two copies of $S_3$.)
Consider now one set containing three alternate edges. The symmetries of hexagon, which move these three alternate edges within themselves will form a group isomorphic to $S_3$ (or better, isomorphic to dihedral group of order 6).
The other set of three alternate edges will give other copy of $S_3$.
Note that, by rotation of $60^0$, one set of alternate edges goes to the other set, hence it is not a member of above two copies of $S_3$. Similarly, rotations by $180^0, 300^0$ are not in above groups; but rotations by $120^0, 240^0, 360^0$ are in above two copies of $S_3$ which you can also see in your answer.