Perhaps I have undersold the 4 and 16 schedule by pointing out the problems. In fact it is quite close to the optimal situation where all pairs occur either once or twice. In total there are 120 pairs, of which only 6 have issues, so the schedule is 95% of the way there. The computer search I used to find it could not do any better.
The twice in a row issue will always be present I fear. The requirement to avoid replication of pairs as much as possible makes this inevitable. For example take the block (6 1 12 4) at machine 1. The second incidence of these 4 players at machine 1, as you might expect, are all in different blocks, so it is more likely than not that one or two of these are in the round before or after. As suggested you are free to reorder the rounds, but I doubt you will be able to achieve much improvement as the problem described will remain.
The 5 and 20 schedule is easier to solve, and I have found an optimal 10 round schedule below, with each player exactly twice at a machine :
( 1 18 5 15) ( 9 8 3 14) ( 4 7 20 11) ( 6 2 17 16) (13 19 12 10)
( 2 19 1 11) (10 9 4 15) ( 5 8 16 12) ( 7 3 18 17) (14 20 13 6)
( 3 20 2 12) ( 6 10 5 11) ( 1 9 17 13) ( 8 4 19 18) (15 16 14 7)
( 4 16 3 13) ( 7 6 1 12) ( 2 10 18 14) ( 9 5 20 19) (11 17 15 8)
( 5 17 4 14) ( 8 7 2 13) ( 3 6 19 15) (10 1 16 20) (12 18 11 9)
( 6 19 14 8) (12 18 20 2) (15 3 10 17) (13 5 11 9) ( 4 1 16 7)
( 7 20 15 9) (13 19 16 3) (11 4 6 18) (14 1 12 10) ( 5 2 17 8)
( 8 16 11 10) (14 20 17 4) (12 5 7 19) (15 2 13 6) ( 1 3 18 9)
( 9 17 12 6) (15 16 18 5) (13 1 8 20) (11 3 14 7) ( 2 4 19 10)
(10 18 13 7) (11 17 19 1) (14 2 9 16) (12 4 15 8) ( 3 5 20 6)
