TITLE: An axiom for collecting proper classes
QUESTION [5 upvotes]: I'm currently working on some universal algebra using proper classes (in MK class theory), and I repeatedly run into situations where I want to collect together some proper classes as the members of a new algebraic structure. 
For example, the construction given here yields a bunch of equivalence classes $\equiv/(m_0,m_1)$ for $(m_0,m_1)\in\mathbb{M}^2$, and if $\mathbb{M}$ is a proper class then each of these equivalence classes is a proper class. Despite this we would like to treat $\mathbb{M}^2/\equiv$ as a group, so to get around the problem we can technically collect together one representative from each equivalence class and be good to go with the class of representatives serving as the new group.
This feels clunky to me from a logical standpoint every time I have to do it, and it makes me wonder if there may be situations where we want to collect up some proper classes which do not admit representatives so simply. Consequently, I would like to add an axiom to MK which allows me to dictate one of two things:

I can collect together proper classes into 'hyper classes' under certain circumstances. This seems like the direct and obvious route at first; it is suggested by Andreas Blass under similar circumstances, however this would require an extension of the formal language of MK to include a relation symbol between proper classes and hyper classes that is not $\in$, and this bothers me.  (this may bother me less as I learn more logic/model theory)
Classes which are definable as equivalence classes using the first order language of sets and a relation $\mathcal{R}$ on a class $\mathbb{B}$ (possibly a proper class) are sets.

The second one is more along the lines of what I want, but this form of it is inconsistent. For example, if we let $\mathfrak{G}(O_n)$ be the Grothendieck ring of the ordinals as defined here for $\omega_1$ (just replace $\omega_1$ with $O_n$ in the construction), we technically need to take a quotient of $\mathfrak{G}(O_n)$ by the equivalence relation $$\equiv=\{(\alpha-\beta,\gamma-\zeta):\alpha+\zeta=\gamma+\beta\}$$ to remove extra 'difference representations' of the same element in $\mathfrak{G}(O_n)$. We now have that $\equiv/(\alpha-\beta)$ is a set for any choice of $\alpha,\beta$ under this new axiom, but $$\equiv/(0-0)=\{\alpha-\alpha:\alpha\in O_n\}$$ is trivially in bijection with $O_n$, a proper class. 

In summary, my question is:

What is a concise axiom that (when added to MK class theory) would allow us to collect together proper classes satisfying certain algebraic formulae as though they were sets? 

If an extension of the language of sets with an additional relation for hyper classes seems the most concise option, I would be open to an argument in favor of that as well.
It also seems from the discussion here that I may want to simply consider collecting these objects together in the meta-theory, but I do not know what pitfalls await me if I try this apparently bold route.

REPLY [6 votes]: Unless you are committed to beginning with MK class theory, which is not conservative over ZFC, I suspect you can get all you want working in Ackermann's set theory as developed by W. Reinhardt in Ackermann's set theory equals ZF, Ann of Math Log 2, pp. 189-249. 
There is a nice overview of the theory by Azriel Levy in The Role of Classes in Set Theory, which appears both as chapter of Foundations of Set Theory (Second Revised Edition), A. Fraenkel, Y. Bar-Hillel and A. Levy, North-Holland Publishing Co. (1973) and as a chapter of Sets and Classes (G.H. Müller ed), North-Holland Publishing Co. (1976).
In Reinhardt's version of Ackermann's theory, which is conservative over ZFC (as well as over NBG with Global Choice), given a class $A$ having the power of $On$ one can form $P(A), PP(A), PPP(A), ...$, where $P(A)$ is the power class of $A$. 
For some reason Ackermann's theory has not received much attention. Perhaps Joel or someone else knowledgable about such matters can explain why this has been the case. 
Edit. I wrote this before I saw Joel's comment.