This question was asked at MO a while ago. Below I quote the relevant part of my answer. But let me add some comments.
In ZFC (proper) classes are not actual objects, we only treat them formally, and they are really just short-cuts for formulas (with parameters), i.e., we identify formulas $\phi(x)$ with the collection of sets that satisfy $\phi$.
This makes talking of relations between classes somewhat awkward. Arturo's nice answer, for example, indicates how one can make sense of, say, a bijection between two classes. A bit more formally, if $X$ is the class of $x$ such that $\phi(x)$ and $Y$ is the class of $y$ such that $\psi(y)$, then a bijection between $X$ and $Y$ is a class $Z$ given by a formula $\rho(x,y)$ such that:
- For any $x,y,z$, if $\rho(x,y)$ and $\rho(x,z)$ then $y=z$.
- Similarly, for any $x,y,z$, if$ \rho(x,y)$ and $\rho(z,y)$, then $x=z$.
- For any $x,y$, $\rho(x,y)$ implies that $\phi(x)$ and $\psi(y)$ holds.
- For any $x$ such that $\phi(x)$ there is a $y$ such that $\rho(x,y)$.
- And, for any $y$ such that $\psi(y)$, there is an $x$ such that $\rho(x,y)$.
This awkwardness also makes it impossible to develop an appropriate theory of classes. For example, the natural question: "Are $X$ and $Y$ pf the same size?" cannot even be asked in ZFC. Of, course, if there is a $\rho$ as above, then we can say that they have the same size, as witnessed by $Z$, and if there is no such $\rho$, outside of ZFC we can say that thy don't have the same size, but all we can do in ZFC is to say, of any given formula $\rho$, that $\rho$ does not define a bijection between $X$ and $Y$.
Of course, equipotency is only an example of what we cannot discuss freely about classes. Thus when interested in proper classes, we move from ZFC to other theories. There are (at least) two natural extensions of ZFC where classes can be treated as formal objects. One is NBG (Von Neumann-Bernays-Goedel). Here, the objects are classes, and sets are classes that belong to other classes. $Z$ is a bijection between $X$ and $Y$ iff $A$ is a function, its domain is $X$ and its range is $Y$ (just as with sets). NBG is usually preferred because it is a conservative extension of ZFC; in a sense, all we have done is to allow references to classes without changing what we mean by "set" in any way. Formally, any theorem of NBG that only mentions sets is a theorem of ZFC. And our interpretation of classes as given by formulas gives us a way of extending any model of ZFC to one of NBG.
The fact that NBG is conservative over ZFC is in a sense a serious limitation, as we limit the notion of class somewhat artificially. When discussing elementary embedding formulations of large cardinal axioms (something very common in modern set theory), for example, the discussion can be carried out in NBG, but not in the smoothest possible fashion. In technical terms, the comprehension axiom in NBG is predicative, but it is hard to justify this when we are allowing proper classes at all.
The most natural extension of ZFC to treat proper classes is Morse-Kelley (MK). Here, comprehension is unrestricted, so it is more natural to combine classes into new ones by usual operations. The cost of this is that MK is not a conservative extension of ZFC. In fact, MK can prove the consistency of NBG (and therefore of ZFC). That being said, to discuss equipotency of classes, MK seems the appropriate framework.
Here is what I said in the MO post mentioned above:
In extensions of set theory where classes are allowed (not just formally as in ZFC, but as actual objects as in MK or GB), sometimes it is suggested to add an axiom (due to Von Neumann, I believe) stating that any two classes are in bijection with one another. Under this axiom, the "cardinality" of a proper class would be ORD, the class of all ordinals. (By the way, by class forcing, given any proper class, one can add a bijection between the class and ORD without adding sets, so this assumption bears no implications for set theory proper.)
Without assuming Von Neumann's axiom, or the axiom of choice, I know of no sensible way of making sense of this notion, as now we could have some proper classes that are "thinner" than others, or even incomparable. Of course, we could study models where this happens (for example, work in ZF, assume there is a strong inaccessible $\kappa$, and consider $V_\kappa$ as the universe of sets, and $Def(V_\kappa)$ in Gödel's sense (or even $V_{\kappa+1}$) as the collection of classes).
Let me expand on this a bit. I argue in ZFC as this is the best known framework of the three mentioned above:
First, class forcing allows us to add a bijection between $V$ and $ORD$ without adding sets. Essentially, what we do is to "thread" through the class of bijections between sets and ordinals. In the resulting extension, we have added no sets, but we have a new class $G$. The resulting structure $(V,G,\in)$ is a model of the strong version of ZFC where we allow $G$ to appear as a predicate in instances of the replacement axiom. Also, any proper class $A$ here (in the sense of "definable by a formula") is also in bijection with the ordinals, and such a bijection is easily definable from $A$ and $G$.
This shows that the assumption that all classes have the same size is completely harmless: No new theorems of ZFC can be proved by adding this assumption.
The model we obtain is not a "natural" model of NBG, though, since G is not definable.
Arturo's suggestion ($V=L$) gives us models where bijections between classes and the ordinals are definable. It has the disadvantage that V=L is quite limiting. As I mentioned in the comments to his answer, there is an alternative: We can assume $G$ definable. This is because, over any model of ZFC, we can force to obtain a new model where $V=HOD$ (on the other hand, once $V\ne L$ we cannot force to have the equality back).
$HOD$ is the class of hereditarily ordinal definable sets. This means that $x\in HOD$ implies that $x\subset HOD$, and that to be in HOD, $x$ must be definable from ordinal parameters. Of course $V=L$ implies $V=HOD$, but $HOD$ is compatible with all known large cardinals, while $L$ is not. Moreover, $HOD$ carries a definable well-ordering (in order type ORD), so if $V=HOD$, then any proper class is in bijection with the ordinals. Moreover, $V=HOD$ is equivalent to the statement that there is a (definable) bijection between $V$ and $ORD$. So, at least in ZFC, this characterizes in a natural way when all proper classes have the same size (and it is really the only way that "sizes of proper classes" can be freely discussed in ZFC).
Finally, it is consistent that $V\ne HOD$, in which case $V$ and $ORD$ have different cardinality in the ZFC sense defined above, and in fact one can arrange that there are incomparable "sizes" of proper classes.
$\newcommand{\Ord}{\operatorname{\mathbf{Ord}}}$Yes, this is true. First, let us prove this if $C$ is a subclass of the ordinals. By transfinite recursion, we can define a function $f:C\to \Ord$ such that for each $c\in C$, $f(c)$ is the least ordinal greater than $f(d)$ for all $d\in C$ such that $d<c$. The image of $f$ is a (not necessarily proper) initial segment of $\Ord$: that is, it is either an ordinal or it is all of $\Ord$. Since $f$ is injective (it is strictly order-preserving), if the image of $f$ were an ordinal then $C$ would be a set by Replacement (using the inverse of $f$). Thus the image of $f$ is all of $\Ord$. But now it is trivial to find a subset of $C$ of cardinality $\alpha$: just take $f^{-1}(\alpha)$ (which is a set by Replacement).
Now let $C$ be an arbitrary proper class. Let $D\subseteq \Ord$ be the class of all ranks of elements of $C$. If $D$ is bounded, then it is contained in some ordinal $\alpha$, which means $C$ is contained in $V_{\alpha}$ and hence is a set. So $D$ must be unbounded, and is thus a proper class. By the previous paragraph, for any cardinal $\alpha$, there exists $S\subset D$ of cardinality $\alpha$. Now use Choice to pick a single element of $C$ of rank $s$ for each $s\in S$. The set of all these elements is then a subset of $C$ of cardinality $\alpha$.
(To be clear, this is a proof you can give for any particular class $C$ defined by some formula in the language of set theory. Of course, ZFC cannot quantify over classes, and so cannot even state this "for all $C$..." at once.)
Best Answer
It is consistent with $\sf ZFC$ that every two proper classes have a bijection between them. That is to say, if $A$ and $B$ are two proper classes then there is a class $C$ which is a class of ordered pairs which is a bijection from $A$ onto $B$.
For example, if we assume that $V=L$, then this is true.
But it is also consistent that this is not the case, and for example there is no bijection between $\sf Ord$, the class of ordinals, and $V$, the class of all sets.
For your motivation, do note that in $\sf ZFC$, or even in class-able set theories like $\sf NBG$ or $\sf MK$ which extend $\sf ZFC$, one cannot talk about classes of proper classes.