The following question appeared in my examination.
Let $G$ be the symmetric group $S_5$ of permutations of five symbols. Consider the set $\mathscr{S}$ of subgroups of $G$ that are isomorphic to the non-cyclic group of order $4$. Let us call these two subgroups $H$ and $K$ belonging to $\mathscr{S}$ as equivalent if they are conjugate (that is, there exists $g \in G$ such that $gHg^{-1}=K$). How many equivalence classes are there in $\mathscr{S}$?
I thought for constructing non cyclic subgroup we would need three elements of order $2$ and order two elements in $S_5$ either would be transposition or it would be product of two transpositions. Because two permutations are conjugate iff they have same cycle representations and thus we would get $2$ equivalence classes. Is this correct? What are other ways to solve this question?
Your reasoning is not correct/complete, but you're on the right track.
You say an order 2 element in $S_5$ "would be transposition or it would be product of two permutations". Presumably you meant "two transpositions" in the sense that their cycle decomposition would be the product of two transpositions.
In any case, this just says that any subgroup of order 4 must consist of elements from two conjugacy classes (together with the identity). It says nothing about the conjugacy classes of such subgroups themselves. You may as well replace the number 4 with the number 2 up to this point--clearly you're missing something.
Edit: Some more discussion was requested. Let's consider $(12)(34)$. Which copies of the non-cyclic group of order 4 can it participate in, if any? Well, $\{e, (12), (34), (12)(34)\}$ comes to mind immediately. Try playing around and seeing if there are any others (hint: there are). Then consider which of them are conjugate (if any). My suggestion is essentially to classify the subgroups first and then ask about their conjugacy afterwards.