In an earlier positing to $\mathcal MO$, it appears that the answer to if the $\sf GCH$ can fail everywhere in every way is to the negative, this is the case in $\sf ZFC$, however it also appears that matters are more free in absence of $\sf AC$. So, here the quenstion is if the opposite direction in provable in $\sf ZF$.
The following question is about if it is true in $\sf ZF$ that for every stage $V_\alpha$ of the cumulative hierarchy there is a cardinal $\kappa > |V_\alpha|$ such that the cardinality of $V_{\alpha+1} \neq \kappa$ in all models of $\sf ZF$?
Formally, this is:
$ \forall M: (M \models \mathsf {ZF}) \implies \\\forall \alpha \in M \, \exists \kappa \in M \big{(} (M \models |V_\alpha| < \kappa) \land \forall N: N \approx M \to N \models|V_{\alpha+1}| \neq \kappa \big{)}$
Where "||" refers to cardinality defined after Scott; $N \approx M$ means that $N$ is a model of $\sf ZF$ that shares the same ordinals and cardinals with $M$, i.e., $\operatorname {Ord}^M = \operatorname {Ord}^N$ and $\operatorname {Card}^M = \operatorname {Card}^N$
In the case of $\sf ZFC$, this seems to be the case, well at least for regular $\alpha$, and the singlular seems to be even more restricted. But, is this the case in $\sf ZF$?
Best Answer
Here's a somewhat trivial answer.
Note that $V_\alpha$, for an infinite $\alpha$, have a particularly nice set of properties which follow from the fact that $|V_\alpha\times V_\alpha|=|V_\alpha|$.
Now, easily, if $\sf AC$ fails, we can find arbitrarily high such cardinals. Simply take $|V_\alpha|+\aleph(V_\alpha)$, where $\aleph(x)$ is the Hartogs number of $x$. By Tarski's lemma, this cardinal is not idemmultiple.
If $\sf AC$ holds, then refer to the previous question you've asked.