Now we need to prove that $gKg^{-1}=K$.So, we try to see the following that $g(12)(34)g^{-1}=g(1)g(2)g(3)g(4) \in K$.Although there are many hints to the problem,I particularly have a problem in one thing - If we take $g(1)g(2)$ as a cycle and $g(3)g(4)$ as a cycle then we try to find the number of possible $2-cycles$in $S_4$.
The possible 2-cycles are of the form $(ab)(cd)$ where the element $a$ has $4$ choices $b$ has $3$ choices .Now the $(ab)$ is the same as $(ba)$ , so we divide $(4×3)/2$ Similarly for the other part we get $(2×1)/2$ Now$(ab)(cd)$ has the same representation as $(cd)(ab)$so we divide the answer by $2$ to get $3$.Hence the number of cycles is .$3$ So they are in $K$.
Now my question is can we choose $g$ in such a way that the conjugate is a $4$ cycle .
An important equality in $\mathcal S_n$ is that if $(a_1 \ a_2 \dots \ a_k)$ is a cycle and $\sigma$ any element of $\mathcal S_n$ then
$$\sigma (a_1 \ a_2 \dots \ a_k) \sigma^{-1} = (\sigma(a_1) \ \sigma(a_2) \dots \ \sigma(a_k))$$
This is used in many areas when dealing with permutations.
Using that, you can prove that the cycle structure of any permutation is preserved by conjugacy.