higher-order reflection In the first-order context, "reflection" of a formula $\varphi(x)$ below $\kappa$ refers to the the following situation: 

There are many ordinals $\alpha<\kappa$ such that for all $a \in V_\alpha$, $V_\alpha \models \varphi(a)$ iff $V_\kappa \models \varphi(a)$.

If $\kappa$ is inaccessible, then this holds for any expansion of $V_\kappa$ in a countable language. "Many" can be taken to mean on a club set.
When we move to higher-order logic, talk of reflection usually shifts to talk of indescribability. A cardinal is $\Pi^m_n$ indescribable if for any $A \subseteq V_\kappa$ and any $\Pi_n$ sentence $\sigma$ in $(m+1)$-order logic with a predicate for $A$, if $(V_\kappa, \in, A) \models \sigma$, then there is $\alpha<\kappa$ such that $(V_\alpha,\in,A\cap V_\alpha) \models \sigma$.  It is a standard fact that if $\kappa$ is measurable, then there is a measure-one set of $\alpha< \kappa$ that are $\Pi^m_n$-indescribable for every $m,n$.  One can also show something stronger: If $\kappa$ is measurable, there is a measure-one set of $\alpha < \kappa$ such that if $A \subseteq V_\alpha$, then there is $\beta < \alpha$ such that $(V_\alpha,\in,A)$ and $(V_\beta,\in,A\cap V_\beta)$ have the same $\omega$-order theory.
Now this is not completely analogous to reflection because we're no longer talking about elementary substructures, but just elementarily equivalent structures, albeit with a common interpretation of a particular predicate.  So my question is, what kind of large cardinal $\kappa$ is needed to get the following statement?

For any $A \subseteq V_\kappa$ and any $n \in \omega$, there are many ordinals $\alpha < \kappa$ such that $(V_\alpha,\in,A \cap V_\alpha) \prec^n (V_\kappa,\in,A)$, where $\prec^n$ means elementary in $(n+1)$-order logic.

It happens at an $\omega$-strong cardinal, but this is clearly not optimal.
 A: It depends on how we define elementarity for the $(n+1)^{th}$-order language of set theory, $\mathcal L^{n}_\in$. For $X\subseteq V_\kappa$, two salient definitions are:
$V_\alpha \prec^n_X V_\kappa$ iff $\forall \vec{x}\in V_\alpha(V_\alpha\vDash \phi(X\cap V_\alpha,\vec{x}) \leftrightarrow V_\kappa\vDash \phi(X,\vec{x}))$, for all $\phi\in \mathcal L^{n}_\in$ with parameters among $X, \vec{x}$. 
$V_\alpha \prec^{*n}_X V_\kappa$ iff $\forall \vec{x}\in V_\alpha\forall \vec{Y}\subseteq V_\alpha(V_\alpha\vDash \phi(X\cap V_\alpha, \vec{Y}, \vec{x}) \leftrightarrow V_\kappa\vDash \phi(X,\vec{Y},\vec{x}))$, for all $\phi\in \mathcal L^{n}_\in$ with parameters among $X,\vec{Y}, \vec{x}$. 
${\bf Theorem}$ ${\bf 1:}$ There are no $\alpha<\kappa$ such that $V_\alpha\prec^{*n} V_\kappa$ for $n\geq 1$. 
${\it Proof.}$ If $Y = V_\alpha$, $V_\alpha\vDash \forall x(x\in Y)$ but not $V_\kappa\vDash \forall x(x\in Y)$. 
${\bf Theorem}$ ${\bf 2:}$ If $\kappa$ is $\Pi^{n}_1$-indescribable for $n\geq 1$, then for any $X\subseteq V_\kappa$ there are unbounded $\alpha<\kappa$ such that $V_\alpha\prec^{n-1}_X V_\kappa$.
${\it Proof.}$ In $n+1^{th}$-order Zermelo set theory we can define a $\Delta^{n}_1$ satisfaction predicate for $\mathcal L^{n-1}_\in$. Denote it $Sat$. Then we can define $Y\subseteq V_\kappa$ such that $V_\kappa$ satisfies:
(*) $\langle \vec{x},\phi\rangle\in Y \leftrightarrow Sat(\phi, X, \vec{x})$, for all $\phi\in \mathcal L^{n-1}_\in$ with parameters among $X, \vec{x}$. 
Clearly, (*) is $\Delta^n_1$ and since $\kappa$ is $\Pi^{n}_1$-indescribable, it follows that there are unbounded $\alpha<\kappa$ which satisfy:
(**) $\langle \vec{x},\phi\rangle\in Y\cap V_\alpha \leftrightarrow Sat(\phi, X\cap V_\alpha, \vec{x})$, for all $\phi\in \mathcal L^{n-1}_\in$ with parameters among $X, \vec{x}$. 
For any such $\alpha$ it is easy to see that $V_\alpha\prec^{n-1}_X V_\kappa$, as required. 
(Note that we can easily strengthen "unbounded" to "stationary" but not "club" - any $\alpha$ for which $V_\alpha\prec^1 V_\kappa$ will be inaccessible.)
${\bf Corollary:}$ $\kappa$ is totally indescribable just in case for any $X\subseteq V_\kappa$ and $n<\omega$ there are unbounded $\alpha$ such that $V_\alpha\prec^n_X V_\kappa$. 
