Say that a logic $\mathcal{L}$ satisfies the weak test property iff for all $\mathfrak{A}\subseteq\mathfrak{B}$ we have $(1)\implies(2)$ below:
For each $\mathcal{L}$-formula $\varphi$ with parameters from $\mathfrak{A}$ we have $$\vert\varphi^\mathfrak{B}\cap\mathfrak{A}^{arity(\varphi)}\vert=\vert\varphi^\mathfrak{B}\vert.$$ (In this case write "$\mathfrak{A}\trianglelefteq_{\mathcal{L}}^{\mathsf{Card}}\mathfrak{B}$.")
$\mathfrak{A}\preccurlyeq_\mathcal{L}\mathfrak{B}$.
This is a massive weakening of the Tarski-Vaught test, which says that we get elementarity merely from $\varphi^\mathfrak{B}\cap\mathfrak{A}^{arity(\varphi)}$ being nonempty whenever $\varphi^\mathfrak{B}$ is nonempty. By contrast, $\mathfrak{A}\trianglelefteq_\mathcal{L}^\mathsf{Card}\mathfrak{B}$ is a highly restrictive hypothesis (and so the corresponding implication is weaker): as long as $\mathcal{L}$ is "reasonable" it immediately implies, for example, that $\vert\mathfrak{A}\vert=\vert\mathfrak{B}\vert$ via the formula $x=x$.
My question is:
Does second-order logic have the weak test property?
Producing interesting instances of $\trianglelefteq_{\mathsf{SOL}}^\mathsf{Card}$, even before trying to also prevent $\preccurlyeq_{\mathsf{SOL}}$, seems very difficult; on the other hand, I see absolutely no reason why $\mathsf{SOL}$should have the weak test property.
In fact there is a whole spectrum of variants of the test property which seem interesting to me. For each class $X$ of cardinals and pair of structures $\mathfrak{A}\subseteq\mathfrak{B}$, say $\mathfrak{A}\trianglelefteq_\mathcal{L}^X\mathfrak{B}$ iff for each $\mathcal{L}$-formula $\varphi$ with parameters from $\mathfrak{A}$ and each $\kappa\in X$ we have $\vert\varphi^\mathfrak{B}\cap\mathfrak{A}^{arity(\varphi)}\vert<\kappa\iff \vert\varphi^\mathfrak{B}\vert<\kappa$; then the weak test property at $X$ is the implication $\trianglelefteq_\mathcal{L}^X\implies \preccurlyeq_\mathcal{L}$. The Tarski-Vaught test itself corresponds to $X=\{1\}$, while the weak test property corresponds to $X=\mathsf{Card}$. If the main question above happens to have a positive answer - which would surprise me quite a bit! - I would be further interested in which $X$s are "sufficient" to ensure $\preccurlyeq_\mathcal{L}$.