No. Let
$$
\alpha=(\kappa^+)^M,
$$
the <successor cardinal> of $\kappa$ computed by $M$. Then $\alpha\in M$, and $M$ regards $\alpha$ as a <cardinal number>. Because $M$ is transitive, $\alpha\subseteq M$, so externally
$$
|\alpha|\leq|M|=\kappa.
$$
On the other hand $\kappa<\alpha$, hence $|\alpha|=\kappa$ in the ambient universe. A corresponding <bijection> belongs to $V_\lambda$ because its <rank of a set>[rank] is below the inaccessible limit $\lambda$. Thus $V_\lambda$ regards $\alpha$ as equinumerous with $\kappa$ and therefore not as a cardinal. This is an instance of <cardinal nonabsoluteness in a small transitive model>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 116 / 3 / b / Solution

Let $M$ be a <transitive set> of cardinality $\kappa$ that models enough set theory and contains $\kappa$. The internal <successor cardinal> $\alpha=(\kappa^+)^M$ is an ordinal in $M$, so transitivity gives $\alpha\subseteq M$ and hence $|\alpha|\leq\kappa$ externally. Although $M$ regards $\alpha$ as a <cardinal number>, an ambient rank containing a bijection between $\kappa$ and $\alpha$ does not. Cardinalhood can therefore fail to be absolute even between transitive models.

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