well, not quite.
suppose (for example) that F contains all 3rd roots of unity, but no primitive 9th root of unity. well all 3rd roots of unity are also 9th roots of unity, if:
a3 = 1 (mod q-1) then:
a6 = (a3)3 = 13 = 1 (mod q-1).
but the subgroup of 9-th roots of unity in F doesn't have order 9, but order 3.
for example F25 has a cyclic multiplicative group of order 24, which therefore has a cyclic subgroup of order 3, but no cyclic subgroup of order 9 since 9 does not divide 24
(if the generator of F* is b, then b8 is a cube root of 1, so b8 is a 9-th root of 1 as well:
(b8)9 = b72 = (b24)3 = 13 = 1).
so all you can say is 93 and 5n-1 have a common factor > 1. this common factor may be 3,31 or 93.
if F5n is supposed to contain ALL 93-rd roots of unity, THEN 93 would have to divide 5n-1.