I don't see any good solutions to your problem, the closest I have would require all 12 teams to be present at the same location in the final round of play.
(1 5) (2 4) | (7 11) (8 10)
(2 6) (3 5) | (8 12) (9 11)
(3 4) (1 6) | (9 10) (7 12)
(1 11) (2 10) | (4 8) (5 7)
(2 12) (3 11) | (5 9) (6 8)
(3 10) (1 12) | (6 7) (4 9)
(7 9) (10 12) | (2 3) (4 6)
(8 7) (11 10) | (3 1) (5 4)
(9 8) (12 11) | (1 2) (6 5)
(1 7) (2 9) | (4 11) (5 10)
(2 8) (3 7) | (5 12) (6 11)
(3 9) (1 8) | (6 10) (4 12)
(7 10) (8 12) | (1 4) (3 5)
(8 11) (9 10) | (2 5) (1 6)
(9 12) (7 11) | (3 6) (2 4)
(1 9) (5 8) (2 11) (4 10)
(2 7) (6 9) (3 12) (5 11)
(3 8) (4 7) (1 10) (6 12)
The friendly matches above all occur in round 5 (columns 2 & 4).
Could you use this schedule?