Let me start by observing that this property is equivalent to a more standard property:
Claim: Let $M\subseteq V$ be a transitive model of $\mathrm{ZFC}$ and let $\mu \in M$ be a cardinal in $V$. The following are equivalent:
- For every $X\in M$, $|X|^V = \mu \iff |X|^M = \mu$.
- $(\mu^+)^V = (\mu^+)^M$.
Proof: Let us assume that $(\mu^+)^V = (\mu^+)^M$. Let $X \in M$. Let us pick a bijection $g\colon X \to \alpha$ in $M$, such that $\alpha$ is an ordinal in $M$. Since $|X|^V = \mu$, $|\alpha|^V = \mu$ and thus $\alpha < (\mu^+)^V$. By our assumption, $\alpha < (\mu^+)^M$ and therefore $|X|^M \leq \mu$. Since $\mu$ is a cardinal in $V$, $|X|^M = \mu$.
For the other direction, let us look at the ordinal $\mu < (\mu^+)^M \leq (\mu^+)^V$. If $\alpha = (\mu^+)^M < (\mu^+)^V$ then $|\alpha|^V=\mu$ and thus, by our assumption, $|\alpha|^M = \mu$ which is impossible. $\square$
As mentioned in the question, if $j\colon V \to M$ is the ultrapower embedding using a normal measure on $\kappa$, then this property fails for $\mu = \kappa^+$ (since $(\kappa^{++})^M < j(\kappa) < (\kappa^{++})^V$). Nevertheless, it holds for $\mu = (2^\kappa)^{+}$, since it and its successor are fixed points of $j$ (namely, $j(\mu) = \mu$ and $j(\mu^+) = \mu^+$). So, this property does not necessarily have consistency strength in the region of strong compactness.
The assumption that $j\colon V\to M$ and $(\kappa^{++})^M = (\kappa^{++})^V$ does have a non-trivial consistency strength (more than measurability). It holds for example, if $\kappa$ is 2-strong and implies that $o^K(\kappa) \geq \kappa^{++}$.