This is not an answer, just following
BillCook's advice and trying out GAP from Sage
sage: g = gap.Group(['(1,2,3,4,5)','(2,3,5,4)']); g
Group( [ (1,2,3,4,5), (2,3,5,4) ] )
sage: g.ConjugacyClasses()
[ ConjugacyClass( Group( [ (1,2,3,4,5), (2,3,5,4) ] ), () ),
ConjugacyClass( Group( [ (1,2,3,4,5), (2,3,5,4) ] ), (2,3,5,4) ),
ConjugacyClass( Group( [ (1,2,3,4,5), (2,3,5,4) ] ), (2,4,5,3) ),
ConjugacyClass( Group( [ (1,2,3,4,5), (2,3,5,4) ] ), (2,5)(3,4) ),
ConjugacyClass( Group( [ (1,2,3,4,5), (2,3,5,4) ] ), (1,2,3,4,5) ) ]
and Sage as well
sage: G = PermutationGroup(['(1,2,3,4,5)','(2,3,5,4)']); G
Permutation Group with generators [(2,3,5,4), (1,2,3,4,5)]
sage: g = G.gens()[0]; g
(2,3,5,4)
sage: g = G.gens()[1]; g
(1,2,3,4,5)
sage: G.order()
20
G.conjugacy_classes_subgroups()
[Subgroup of (Permutation Group with generators [(2,3,5,4), (1,2,3,4,5)]) generated by [()], Subgroup of (Permutation Group with generators [(2,3,5,4), (1,2,3,4,5)]) generated by [(2,5)(3,4)], Subgroup of (Permutation Group with generators [(2,3,5,4), (1,2,3,4,5)]) generated by [(2,3,5,4)], Subgroup of (Permutation Group with generators [(2,3,5,4), (1,2,3,4,5)]) generated by (1,2,3,4,5)], Subgroup of (Permutation Group with generators [(2,3,5,4), (1,2,3,4,5)]) generated by [(2,5)(3,4), (1,5,4,3,2)], Subgroup of (Permutation group with generators [(2,3,5,4),(1,2,3,4,5)]) generated by [(2,3,5,4), (2,5)(3,4), (1,5,4,3,2)]]
# this is an ugly hack:
for x in G.conjugacy_classes_subgroups():
s=str(x).replace('Subgroup of (Permutation Group with generators [(2,3,5,4), (1,2,3,4,5)]) generated by ','')
print s.replace('[','<').replace(']','>')
<()>
<(2,5)(3,4)>
<(2,3,5,4)>
<(1,2,3,4,5)>
<(2,5)(3,4), (1,5,4,3,2)>
<(2,3,5,4), (2,5)(3,4), (1,5,4,3,2)>
to show the results.
I think there is a slick trick to know but I don't remember it.