This is a try to salvage the attempt written in the posting: "Can we derive a large cardinal axiom by a principle of isomorphism between cardinals and ordinals?"
Here we change the base theory to Ackermann's set theory +Choice + Foundation over all classes
To restate the principle: for any property $\psi$ that is parameter free definable in the langauge of Ackermann set theory that doesn't use the symbol $V$ [standing for the class of all sets], such that for every set ordinal $x$ there is class cardinal $y$ that satisfy $\psi$ such that $x$ is strictly smaller than $y$, then the class of all sets (i.e. elements of $V$) that are cardinals, is isomorphic to the class of all ordinals that are sets.
Note: to avoid confusion, the terms 'cardinal' and "ordinal" are defined in the usual manner following Von Neumann's, but they are not restricted to elements of $V$.
Formally this is:
Cardinals to Ordinals isomprhism: if $\psi(Y)$ is a formula that doesn't use the symbol $V$, in which only the symbol $``Y"$ appear free [and only free], and the symbol $``x"$ never occurring, and $\psi(x)$ is the formula obtained from $\psi(Y)$ by merely replacing each occurrence of the symbol $``Y"$ by an occurrence of the symbol $``x"$, then:
$$\forall x \in V (x \text{ is an ordinal } \to \exists Y[\psi(Y) \wedge Y \text{ is a cardinal } \wedge x < Y]) \\ \to \{x \in V | x \text{ is a cardinal } \wedge \psi(x) \} \text{ is order isomorphic to } \{x \in V| x \text{ is an ordinal}\} $$
is an axiom.
We also add to this theory the axiom that $V$ is of regular cardinality.
The idea is that "regular" is definable in a parameter free manner without using $V$, and so it would fulfill the antecedent of that principle, so by this principle the class of all regular cardinals in $V$ must be order isomorphic to the class of all ordinals in $V$, this is stronger than the usual Ackermann set theory, and also stronger than both $ZF$ and $MK$.
So the question is about the consistency strength of this theory?