Divisors:
1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120 

1,120: trivial

60: must be normal (Index 2) and therefore union of classes. There is only
one such possibility, namely A5.

2: Cyclic (1,2) [10=5 choose 2] and (1,2)(3,4) [15=5 choose 2*3 choose 2 /2]

5:  Cyclic (1,2,3,4,5), Sylow subgroup, (Only this class, number is >1,
     only divisor of 120 that is congruent 1 mod 5 is 6)
     [6 subgroups]
     Therefore: Let S be the normalizer of a 5-Sylow subgroup: [G:S]=6, so
     |S|=20. So there are 6 subgroups of order 20, which are the Sylow
     normalizers. 

20:  Vice versa. If |H|=20, the number of 5-sylow subgroups (congruence and
     dividing) must be 1, so the *only groups of order 20 are the sylow
     normalizers.
     <(1,2,3,4,5),(2,3,5,4)>.

10: Ditto, a subgroup of order 10 must have a normal 5-Sylow subgroup, so
    they must lie in the 5-Sylow normalizers, and are normalized by these.
    The 5-Sylow normalizers have only one subgroup
    of order 10: [6 subgroups of order 10]

15: would have a normal 5-Sylow subgroup, but then the 5-Sylow nomalizer
would have to have an order divisible by 3, which it does not -- no such
subgroups.

40: would have a normal 5-Sylow subgroup, but then the 5-Sylow nomalizer
would have to have order 40, which it does not -- no such subgroups.

8: 2-Sylow subgroup, all conjugate. We know that [S_5:S_4]=5 is odd, so a
2-Sylow of S4 is also 2-Sylow of S5. D8=<(1,2,3,4),(1,3)> fits the bill.
Number of conjugates: 1,3,5 or 15 (odd divisors of 120). We know that we can
choose the 4 points in 5 ways, and for each choice there are 3 different
D8's: [15 subgroups]
We therefore know that the 2-Sylow subgroup is its own normalizer.

4:  Must be conjugate to a subgroups of the 2-Sylow subgroup. Therefore the
following three are all possibilities:
Cyclic (1,2,3,4) [15=5 * 3: choose the point off, then there are 3 subgroups
each]
   not cyclic: (1,2),(3,4) [15, ditto ]
     and (1,2)(3,4),(1,3)(2,4) [5 copies: choose the point off]

3: Cyclic (1,2,3) [5 choose 3=10 subgroups as 1,2,3 and 1,3,2 lie in the same
group] Also Sylow, thus the 3-Sylow normalizer must have index 10, order 12.

6: A group of order 6 must have a normal 3-Sylow subgroup, and therefore lie
in a 3-Sylow normalizer, which is (conjugate to) <(1,2,3),(1,2),(4,5)>. We
can find <(1,2,3)(4,5)> (cyclic) [5 choose 3=10 groups], 
<(1,2,3),(1,2)> (S3 [5 choose 3= 10 copies]) and
<(1,2,3),(1,2)(4,5)> (isomorphic S3) [5 choose 3=10 copies]

30: can have 1 or 10 3-Sylow subgroups. If 1, it would be in the Sylow
normalizer, contradiction as before. If it is 10, it contains all 3-Sylow
subgroups, but <(1,2,3),(3,4,5)> has order 60, Contradiction.

12: If it has a normal 3-Sylow subgroup, it must be amongst the  3-Sylow
normalizers <(1,2,3),(1,2),(4,5)> [10].
If not, it contains 4 3-Sylow subgroups, i.e. 8 elements of order 3. Thart
leaves space only for 3 elements of order 2, the 2-Sylow subgroup (or order
4) therefore must be normal, and its normalizer has index at most 10, i.e.
it has at most 10 conjugates. This leaves <(1,2)(3,4),(1,3)(2,4)>, whose
normalizer is S4. (its normal in S4, and S4 has the right order). Inspecting
S4, we find it has only one subgroup of order 12, namely
A4=<(1,2,3),(2,3,4)> (see Homework) with [5, same as the number of S4's]
conjugates.

24: We know there are [5] copies of S4. We now want to show that is all:
(There are 15 groups of order 24, so we can't that easily determine the
structure alone from the order.)
Its 2-Sylow subgroup must be a 2-Sylow of S5, so WLOG its 2-Sylow is D8.
Now consider the orbit of 1 under this subgroup. It must have length at
least 4 (that's under D8), but the orbit length must divide 24. So it cannot
be 5. But that means that the subgroup fixes one point and thus is S4.