Let $\mathsf{ZF^-_2}$ be a second-order ZF without Powerset, and the second-order Collection in place of Replacement. It is easy to see that every transitive model of $\mathsf{ZF^-_2}$ is closed under subsets of its elements.
Moreover, we can show in $\mathsf{ZFC}$ that every transitive model of $\mathsf{ZF^-_2}$ is $H_\kappa$ for some regular $\kappa$:
Proof. Let $\kappa=\sup\{|x|:x\in A\}$, then $A\subseteq H_\kappa$. By second-order Replacement with the induction, we have $\alpha\in H_\kappa$ if $\alpha<\kappa$. Applying second-order separation, we can see that $A$ and $H_\kappa$ agrees on its sets of ordinals. Hence $A=H_\kappa$.
My question is:
Is it consistent with $\mathsf{ZF}$ that $A$ is a transitive model of $\mathsf{ZF^-_2}$, but $A\neq H_\kappa$ for any cardinal $\kappa$?
Since the notion $H_\kappa$ is not unique in the choiceless setting, the answer could depend on the formulation of $H_\kappa$. (I know there are at least two definitions of it: one of them is that of Aspero and Karagila, and the other one is that of Lubarsky and Rathjen.)
Note that, Lubarsky and Rathjen proved the following: if $A$ is a transitive set such that $2\in A$ and $A$ satisfies Union and second-order Replacement, then the ordinal $$o(A) = \min\{\alpha\mid\alpha\notin A\}$$ is regular. (Lemma 2.4 of Lubarsky and Rathjen.) Hence the following strategy does not work: construct a model that thinks every cardinal is singular but there is a transitive model of $\mathsf{ZF^-_2}$.
I would appreciate any help!
