Given elements $a$ and $b$ in a group $G$, $a$ is said to be conjugate to $b$, if there is some $c \in G$ satisfying $c^{-1}ac = b$.
Now, conjugacy is an equivalence relation on elements of a group, because :
$a = eae=e^{-1}ae$, where $e$ is the identity.
If $a = c^{-1}bc$, then $b = (c^{-1})^{-1} a c^{-1}$.
If $a = d^{-1}bd$ and $b = f^{-1}cf$ then $a = (fd)^{-1}c(fd)$.
Hence, the equivalence class of $a$ i.e. the set of elements which are conjugate to $a$, are called the conjugates of $a$.
For $S_n$, we have the cycle notation :for example $(abc)$ is the permutation taking $a \to b, b\to c, c\to a$. Every element can be written as a product of disjoint cycles.
Now, let $h$ be a permutation in $S_n$, and $g = (a_1...a_i)(b_1...b_j)...(t_1...t_l)$ be some element of $S_n$ written as a product of disjoint cycles.
Then, turns out $h^{-1}gh $ has a cycle notation : $(h(a_1) ... h(a_i))(h(b_1)...h(b_j))...(h(t_1)...h(t_l))$ as a product of disjoint cycles. Therefore, $h^{-1}gh$ retains the disjoint cycle structure of $g$ : the same number of cycles, and each cycle has the same number of elements.
Conversely, given any permutation $f$ with the same disjoint cycle structure as $g$, then a $h$ will exist so that $h^{-1}gh = f$. Write down $f$ like how I wrote down $g$ and see how $h$ can be described.
From the above paragraphs follows :
Two permutations are conjugate if and only if they have the same cycle structure.
Hence, all you need to do, is to list permutations, which have the same cycle structure as $(13)(24)$, which means that it is of the form $(ab)(cd)$, where $\{ a,b,c,d\} = \{ 1,2,3,4\}$.
So, in this case, you would get permutations that look like this:
$(12)(34)$
$(13)(24)$
$(14)(23)$
The above three are conjugate, and no other permutation is conjugate to any of these.
