Actually, no. The number depends on the position at the table—rotations are distinct.

For 3 couples:

Mrs A must sit between Mr B and Mr C. Mr B can be on her left or her right, 2 possibilities. After those two seats are chosen, everyone else’s seat is determined.

But there are 6 places Mrs A might sit. Hence, for 3 couples there are 12=2x6 possible seatings, including rotations.

I don't believe so. I worked out the permutations for n = 3 and, accounting for rotations, you only get 2: [0, 3, 4, 1, 2, 5] [0, 5, 2, 1, 4, 3]. Of course, you get the expected answer of 12 if you multiply by 6.

It's doesn't really matter, the difference is trivial.

you can just divide out the rotations if your dinner party plans are rotation invariant.