The answer is yes.
KM commonly taken to include the axiom of choice and furthermore the axiom of global choice, and this makes things very easy, since we can just let $C(X)$ be the smallest ordinal that is equinumerous with $X$, if $X$ is a set, and otherwise some non-cardinal default value for the proper classes, since these are all equinumerous under global choice.
If you drop the axiom of choice, then it is still possible to get a solution to the cardinal-assignment problem for sets only, that is, when $X$ is a set, by defining $C(X)$ to be the set of minimal-rank sets that are equinumerous with $X$. This is an equinumerosity invariant as desired. (And in fact, this application is the original use of Scott's trick, solved by Scott as an undergraduate at Berkeley in response to a question posed to the class by Tarski.)
Without global choice, it is not in general possible to find a solution to the cardinal assignment problem, since there could simply be more equinumerosity classes amongst the proper classes than there are sets. For example, if $\kappa$ is an inaccessible cardinal and $2^\kappa\geq\aleph_{\kappa^+}$, then $V_{\kappa+1}$ is a model of KM, but there are too many sizes of classes to find an invariant amongst the sets.
Meanwhile, there is a very general problem here regarding Fregean abstraction, where one wants to assign abstractions with respect to general equivalence relations. The problem has many variations, depending on whether the field of the relation is just sets or classes, and whether the equivalence relation is first-order definable or second-order definable and so forth.
I mounted a very general analysis in my paper:
- Joel David Hamkins, Fregean abstraction in Zermelo-Fraenkel set theory: a deflationary account, arxiv:2209.07845.
The paper has several theorems providing solutions to the Fregean abstaction problem in many variations.