Let $X, Y$ be two proper classes. We say that the power of $X$ is equinumerous to $Y$ iff there is a relation $R$ with $domain(R) = Y$ and $range(R) = X$ such that (i) $\forall Z \subseteq X \exists y \in Y R_y = Z$ (where $R_y$ is $\{x : <y ,x> \in R\}$), and (ii) $\forall y_1 , y_2 \in Y \ (y_1 \neq y_2 \rightarrow R_{y_1} \neq R_{y_2})$. (So R codes a ''bijection'' from Y to the ''powerclass'' of X)
My question is: does KM - Global Choice prove the following statement (*) (which basically says that $V$ cannot be reached by taking ''power'' of any class smaller than it)?
(*) If there is no injection from $V$ to $X$, then the power of $X$ is not equinumerous to $V$.
(Since Global Choice implies that all proper classes are equinumerous, this is not an interesting statement given GC.)
No, this is not the case.
The counterargument is exactly along the lines Asaf suggested. If $M\models\mathsf{ZFC}$ then there is a definable-over-$M$ class surjection from the class of sets of ordinals to $M$ itself. Basically, with a set of ordinals we can "code" arbitrary initial segments of the cumulative hierarchy. In detail, for a set $u$ say that a code for $u$ is a triple $(\alpha,R,\beta)$ where $\alpha$ is some ordinal, $R$ is a binary relation on $\alpha$ such that $(\alpha;R)\cong V_\gamma$ for some $\gamma$ with $u\in V_\gamma$, and the image of $\beta$ under the unique isomorphism $(\alpha;R)\cong V_\gamma$ is $u$ itself. By appropriate tuple-juggling, we can replace codes with sets of ordinals.
The point is that if global choice fails, there is no injection from $V$ to $Ord$. However, running the above idea with $M=V$ shows - informally speaking - that the powerclass of $Ord$ is at least as big as $V$.
Note that there's some subtlety around the notion of "powerclass" here: are we talking about the collection of all subclasses of a given class, or the collection of all subsets of a given class? Call these the "big powerclass" and "small powerclass" respectively. The argument above gives a strong negative answer to your question: although $V$ doesn't inject into $Ord$, already the small powerclass of $Ord$ is as big as $V$.