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Clearly, $p_0$ meets everyone once and the other $p_i$ are symmetric; so it suffices to check that $p_1$ meets everyone.

Here is the list of partners for $p_1$, as everyone but $p_0$ moves counterclockwise each round: $p_0, p_, p_, \ldots, p_4, p_2, p_, p_, p_, \dots , p_5, p_3$. I changed the formatting in the answer from "vertical" (which wasn't working) to "horizontal", but feel free to revert.

Within each group, the ones whose bit after the prefix is 0 are evens and the others odds. (Each person in the top row "meets" the corresponding person just below in the bottom row.) 1) (#1:4 #2:5 #3:6 #4:1 #5:2 #6:3) I tried to show the stages above. Draw it out: 1) draw six circles (a network graph), then draw a line w/ 1,2,3 on one side, 4,5,6 on the other. For each couplet (sitting across the line), make a line on the graph connecting them. Round 2, rotate all the numbers around the line, except 1 (it's in a box). For any group of 2n people, there will be 2n-1 rounds, with everyone meeting everyone else once.

Although, the best way is still to work everything out in advance, write it down on cards and have everyone take a card on arrival. If you have an odd number of people, just add an imaginary person to the mix, with the result that each person has one round off from meeting anyone.

The first half is easy, you just have the odds stay in their seats and rotate evens. Regardless, i think the long narrow solution below works.) Can you clarify the requirements you'd like to impose exactly? you may worry less about two people swapping seats than about half the people getting up and moving elsewhere).