Hey the number of conjugates is (nC2)*((n-2)C2)/2 ? Is that right?
Now I am trying to compute the general form of elements that commute with (1,2)(3,4). I am not sure how to go about that..
For example, let n=7. You can arrange your cycle type (x1 x2)(x3 x4)(x5)(x6)(x7) in 7! ways. You need to divide some overcount numbers. You see that (1,2)(3,4), (1,2)(4,3), (2,1)(3,4), (2,1)(4,3) are all the same , corresponding 2^2 factor. You can also see that each (x1 x2)(x3 x4) is the same with (x3 x4)(x1 x2), corresponding 2! factor. Now the remaining permutation involves (x5)(x6)(x7), corresponding 3! factor. For n=7, the number is 7!/(2^2 *2! *3!).