Is there an isomorphism class which is a set? I'm studying Set Theory in my own, with Goldrei's textbook. The chapter I'm reading is on order-isomorphism and well-ordering. One exercise asks (i) to argue that, in general, a collection of well-ordered sets order-isomorphic to a given well-ordered set is a proper class (rather than a set). The proof, IMHO, is fairly straightforward and applicable to any isomorphism class. Then the same exercises asks (ii) if there is a well-ordered set, X say, s.t. the collection of all well-ordered set, order-isomorphic to X, is a set (rather than a proper class). I stumbled on that one.
 A: The class of the empty well-order is the only such class that is also a set. For every other class $[(X,\le)]$ the class-function sending $(Y,\preceq)\mapsto Y$ would be onto the class of sets that are in bijection with $X$, which is proper as soon as $X\ne\emptyset$.
A: Let’s assume $X\neq\emptyset$ (else the property is trivially true). First note that you are looking at classes of order-isomorphic ordered sets, so pairs $(A,\leq)$ so that all for all such pairs exists a bijection that leaves the order intact. Let’s call this class $Z$. On the other hand if there is a bijection between two sets $A,B$ then this will transform any ordering on $A$ into an ordering on $B$.
Thus by projecting on the first coordinate we get $|X|$ where $|X|$ denotes the class of cardinality.
So if $Z$ is a set, then also $|X|$ is a set.
Now, can this be? Let’s differentiate two cases: If $|X|=n>0\in\mathbb N$ then $B:=\{0,1,\ldots,n-2,|X|\}$ would be a set of that cardinality. On the other hand, if $|X|$ is infinite then $B=X\cup\{|A|\}$ has the same cardinality as $X$.
So in both cases we’d find some $B\in|X|$ so that $|X|\in B$. By ZF this is not possible.
Thus this is not in fact a set.
EDIT: On the ZF part. Set $C=\{B,|X|\}$. Then $C,B$ are not disjoint, for both contain $|X|$, and $C,|X|$ are not disjoint, for both contain $B$. Thus this violates the axiom of regularity.
