well, you know that s has to map the set {1,2,4} to the set {1,2,4} and the set {3,5} to the set {3,5}.

this gives us TWO choices for s on 3 and 5, either:

s(3) = 3, s(5) = 5 -OR-

s(3) = 5, s(5) = 3.

the first set ({1,2,4}) is a little trickier. we have 3 choices for s(1): s(1) = 1, s(1) = 2, or s(1) = 4.

if s(1) = 1, then s(4) must be 2, and thus s(2) = 4.

if s(1) = 2, then s(4) must be 4, and s(2) = 1. for example, one can verify that if s = (1 2), that:

(1 2)[(1 4 2)(3 5)](1 2) = [(1 2)(1 4 2)][(1 2)(3 5)] (since (1 2) and (3 5) commute)

= [(1 2)(1 4 2)(1 2)](3 5) = (2 4 1)(3 5) = (1 2 4)(3 5) (since (2 4 1) = (1 2 4)).

if s(1) = 4, then s(4) = 1, and s(2) = 2.

so s is completely determined by s(1), and s(3). we have 3 choices for s(1), and 2 choices for s(3), giving 6 possible s:

s = (2 4)

s = (2 4)(3 5)

s = (1 2)

s = (1 2)(3 5)

s = (1 4)

s = (1 4)(3 5) is a complete list.

keep in mind, however, that this is a somewhat artificial example. if we are dealing with S9, for example, s might map any element of {6,7,8,9} to any other (which wouldn't affect the conjugate, since (1 4 2)(3 5) doesn't affect 6,7,8 or 9, so as long as s takes an element of {6,7,8,9} to another element of {6,7,8,9}, s^-1 just takes it back). so in that case, we get 24 permutations for each of the s i listed above. so there's no "general formula", the answer depends on which symmetric group you're conjugating IN, and not just the conjugates themselves.

keep in mind, as well, that symmetric groups Sn (except for very small values of n), tend to be quite large, and that finding all elements that conjugate a to b, can be a tedious task. usually, finding one that works is "good enough".