Equivalence relation of transposition has equivalence groups of all same size. Let $G$ be a transitive subgroup of the permutation group $S_d.$ Now define an equivalence relation on $S = \{1, \ldots, d\}$ by $i \sim j$ iff the transposition $(ij)$ exists in $G.$ How do I show that all the equivalence classes have the same size by the fact that $G$ is transitive? Suppose $A_1$ and $A_2$ are two equivalence classes. I can perhaps take a permutation of $G, \sigma, $ such that $\sigma(x) = y$ for $x \in A_1$ and $y \in A_2.$ However, I don't think it is true that $\sigma(A_1) = A_2.$ I'm not sure how to see this. Any hints? 

Let $C_i=\{(i, j) \in G\} $. Let $g\in G$. Can you prove that
$$C_{g(i)} =\{g(i, j)g^{-1}\mid  (i, j) \in C_i\}?$$
Then you have to prove that the size of $A_i$ is the size of $C_i$ plus $1$. By transitivity, you have bijections between the sets of transpositions, hence the result. 
