We know that every equivalence relation, induced by a partition on a set for example $X$ , make some equivalence classes. Now, if a group $G$ acts on $X$ then the associated equivalence classes are exactly the orbits of $X$ under the action of $G$. In fact, if $x\in X$ then we know $[x]$ in which $[x]=\{x^g\mid g\in G\}$ as an equivalence classes for $X$. Here, I am facing an example in permutation groups as following
Let $G=\operatorname{Aut}(\mathbb Q,\le)$ be the group of permutations of $\mathbb Q$ which preserves the usual ordering $\le$ . Also, suppose that $G$ acts on $\mathbb Q^2$ transitively, so we have two orbits of this action; $\Delta:=\{(\alpha,\beta)\in \mathbb Q^2\mid \alpha<\beta\}$ and $\Delta^*:=\{(\beta,\alpha)\mid (\alpha,\beta)\in ∆\}$ clearly.
I am misunderstanding of what would be that representing $x$ above for one of these two orbits in the example? Could every element of $\Delta$ be taken as a representing element to show the equivalance class? Indeed, I am baffling of the way we denote $\mathbb Z_p=\{[0],[1],[2],\dots,[p-1]\}$ every time we need it. Thanks for sharing the thoughts.