Is there a single well-ordered set whose discrete space equals to its order topology?

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Let $(X,\leq_X )$ and $(Y,\leq_Y)$ be two well-ordered sets, such that the discrete space of $X$ is the same as the order topology on $(X,\leq_X)$ and the same goes for $Y$. I'd like to prove that $(X,\leq_X)\cong (Y,\leq_Y) $. It seems correct, but I'm having trouble on where to start from.

Another thing is that I also need to prove that there is an infinity of total ordered and countable sets $(Z_i,\leq_{Z_i})$ such that the discrete space of each $Z_i$ equals the order topology of $(Z_i,\leq_{Z_i})$, and no pair of them is isomorphic.

Now, this got me even a bit more confused, because I'm not sure it's even correct. Is it true? And if so, is the reason for this difference is because at first I was talking about well-ordered sets and now I'm talking about total orders? In any case, I feel helpless as to where to start in both the cases.