For general equivalence relations, not those arising from permutation group actions, even finding lexicographically least is still "too" general. Finding the lexicographically smallest element in an equivalence class can be $NP$-hard (in fact, $P^{NP}$-hard) - even if the relationship has a polynomial-time canonical form [1].
However, for permutation group orbit problems as you describe, deciding whether two points lie in the same orbit is not likely to be $NP$-hard: it is in $NP \cap coAM$, and hence not $NP$-hard unless the polynomial hierarchy collapses to the second level.
A canonical form for graph isomorphism is also a special case of the second problem you state. The best known canonical form for graph isomorphism runs in time $2^{\tilde{O}(\sqrt{n})}$ [2].
Since you said in the comments that any canonical form will do, you might also be interested in my paper with Lance Fortnow [3]: in its currently generality, I think your question is related to our results. We show that if every equivalence relation decidable in $P$ has a canonical form in $P$, then "bad" consequences result, such as $NP = UP = RP$, which in particular implies that the polynomial hierarchy collapses down to $BPP$. On the other hand, the equivalence relations you're interested in may not be in $P$, but this result suggests that even if it lies in a higher complexity class other hard problems may still stand in your way.
So I think if you want some better upper bounds you really need the problem to be more specific.
[1] Andreas Blass and Yuri Gurevich. Equivalence relations, invariants, and normal forms. SIAM J. Comput. 13:4 (1984), 24-42.
[2] László Babai and Eugene M. Luks. Canonical labelings of graphs. STOC 1983, 171-183.
[3] Lance Fortnow and Joshua A. Grochow. Complexity classes of equivalence problems revisited. Inform. and Comput. 209:4 (2011), 748-763. Also available as arXiv:0907.4775v2.