For n = 2, a trivial solution is to have one die with three 1s and three 3s and one die with six 2s.
In general there are 6^n different outcomes and n! different orderings, so a necessary condition is that 6^n is a multiple of n! That means any n ≥ 5 won’t work.
For equal chance of all orderings, yes. "Equal chance of going first" is more flexible, though I expect at some point we hit a limit in expressiveness (at least) if we're still using 6 sided dice.
I don't know whether any of those can be non-transitive, though. I'd guess that it's sometimes possible but it's only a guess.
In general there are 6^n different outcomes and n! different orderings, so a necessary condition is that 6^n is a multiple of n! That means any n ≥ 5 won’t work.