In the 30-object, 5-bucket case, as veefu points out, only one round is possible, since the six objects in bucket 1 have to go into 6 different buckets in round 2, but there aren't 6 buckets for them to go into.
90 objects with 15 buckets, object number 1 must be with 5 new objects every round, there are 89 objects other than object 1, so at most 17 rounds are possible ($17\times5\le89\lt18\times5$). Whether 17 rounds are actually possible is another (and generally much harder) question. Similarly, with 120 objects in 20 buckets, you certainly can't have more than 23 rounds, but it's hard to say whether 23 rounds are possible.
Fortunately, a lot of people have put a lot of work into solving this sort of question, and there are tabulations available in journals and on websites. Whether they go all the way up to 120 objects in 20 buckets, I wouldn't know. Perhaps a good search term would be "combinatorial designs".
EDIT: Another search term that might turn up something is "social golfer problem". This concerns a number of people who want to play as many rounds of golf as possible, going out in groups of 4, no two players being in the same group of four for more than one round.