Combinatorics – Schedule for 2 Tables of 6 People to Share Table Equally

Question

The problem
There are 2 tables seating 6 people each. With 12 people, how many arrangements (with all 12 people seated) are necessary so that every pair shares a table for the same number of arrangements?

Example
Here's an example with 4 arrangements:

Arrangement 1: 1,2,3,4,5,6 and 7,8,9,10,11,12
Arrangement 2: 1,2,3,4,11,12 and 7,8,9,10,5,6
Arrangement 3: 1,2,9,10,5,6 and 7,8,3,4,11,12
Arrangement 4: 7,8,3,4,5,6 and 1,2,9,10,11,12

1 and 2 share a table 4 times.
1 and 3 share a table 2 times.
1 and 4 share a table 2 times.
…
11 and 12 share a table 4 times.

This is an invalid answer, because not all pairs share a table the same number of times.

My reasoning so far
In a single arrangement, a given person shares a table with 5 people out of 11 (other people). So, that person must sit with every other person 5 out of 11 times. So the answer must be a multiple of 11.

There are $\frac{\binom{12}{6}}{2}= 462$ unique arrangements, so that's an upper bound. (Division by two because the two tables are interchangeable).

(I'm also curious about the same problem, but with 2 tables of $n$ people)

Answer 1

$1,2,3,5,8,12$
$1,3,4,6,9,2$
$1,4,5,7,10,3$
$1,5,6,8,11,4$
$1,6,7,9,12,5$
$1,7,8,10,2,6$
$1,8,9,11,3,7$
$1,9,10,12,4,8$
$1,10,11,2,5,9$
$1,11,12,3,6,10$
$1,12,2,4,7,11$.