Is there a single well-ordered set whose discrete space equals to its order topology? 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.
 A: For the first part, as Henno Brandsma mentions in his comment, (up to order-isomorphism) the only well-ordered sets which are discrete under the order topology are the finite ones (i.e., $0 < 1 < \cdots < n-1$ for some natural number $n$), and $\mathbb{N}$ itself.  That each of these are discrete under the order topology is fairly easy to check.  
We can salvage the question by additionally assuming that $X$ and $Y$ are both infinite.  In this case, what we need to show is that if $\langle X , \leq \rangle$ is an infinite well-ordered set which is discrete in the order topology, then $\langle X , \leq \rangle \cong \langle \mathbb{N} , \leq \rangle$.
Outline: If $\langle X , \leq \rangle$ is an infinite well-ordered set which is not (order-)isomorphic to $\langle \mathbb{N} , \leq \rangle$, then $\langle \mathbb{N} , \leq \rangle$ is isomorphic to an initial segment of $\langle X , \leq \rangle$; that is, there is an injective order-preserving function $f : \mathbb{N} \to X$ such that the image of $f$ is an initial segment of $X$.  Now this mapping cannot be onto, and so consider the least element $x$ of $X \setminus \{ f(n) : n \in \mathbb{N} \}$.  Show that $\{ x \}$ is not open.

For the second part, here's a rather small
Hint: The total order consisting of two copies of $\mathbb{Z}$ one after the other is not (order-)isomorphic to $\mathbb{Z}$.
A: For the first part, you first need that |X| = |Y|. Then there exists f: X -> Y which is bijective. Since the order topologies on each is really just the discrete topology, f is open and continuous. So, f is a homeomorphism
