Let $X$ be a non-empty arbitrary (even infinite) set. It is well known that there exists at least a well-ordering on $X$. Actually, there are different well-orderings on $X$. Many of them turn out to be isomorphic. So my question is: how many non-isomorphic well-orderings is it possible to construct on an arbitrary non-empty ground set $X$? Something is said here:
Number of well-ordering relations on a well-orderable infinite set $A$?
but it would be interesting for me to have a complete answer to my question, if possible.
We work in ZFC, so every set can be well-ordered.
If $A$ is finite, there is only one well-order of it. This is not hard to see.
If $A$ is infinite and of size $\kappa$, then each non-isomorphic well-ordering corresponds to a unique ordinal of size $\kappa$. There are exactly $\kappa^+$ ordinals of size $\kappa$, and so there are $\kappa^+$ non-isomorphic well-orderings of $A$. As Sassatelli points out, this is what Hartogs' Lemma says.