Let $\kappa$ and $\lambda$ be cardinals. A thin $(\kappa,\lambda)$-list is a function $L:[\lambda]^{<\kappa}\longrightarrow [\lambda]^{<\kappa}$ such that for all $x\in[\lambda]^{<\kappa}$, $L(x)\subseteq x$ and $\{L(y)\;|\;y\subseteq x\}$ has cardinality $<\kappa$.

Say that $(\kappa,\lambda)$-STP holds iff whenever $L$ is a thin $(\kappa,\lambda)$-list, there is some $b\subseteq\lambda$ such that $\{x\;|\;x\cap b=L(x)\}$ is stationary and that $\kappa$ satisfies the super tree property iff $(\kappa,\lambda)$-STP holds for all $\lambda\geq\kappa$.

Say that $(\kappa,\lambda)$-SSTP holds iff for every sequence $(L_{\alpha})_{\alpha\in\mu}$ (where $\mu<\kappa$), if every $L_{\alpha}$ is a thin $(\kappa,\lambda)$-list, there is a sequence $(b_{\alpha})_{\alpha\in\mu}$ such that $\{x\;|\;\forall\alpha\in\mu(b_{\alpha}\cap x=L_{\alpha}(x))\}$ is stationary. Say that $\kappa$ satisfies the simultaneous super tree property iff $(\kappa,\lambda)$-SSTP holds for all $\lambda\geq\kappa$.

Let $\kappa$ be supercompact, $\lambda\geq\kappa$ and $U$ a normal $\kappa$-complete ultrafilter on $[\lambda]^{<\kappa}$. For a thin $(\kappa,\lambda)$-List $L$ one can show that there is some $b$ such that $\{x\;|\;x\cap b=L(x)\}\in U$, therefore, using $\kappa$-completeness, we have that supercompact cardinals satisfy the simultaneous super tree property, so the simultaneous super tree property is consistent, at least modulo the existence of a supercompact cardinal (and I suspect $\omega_2$ has the simultaneous super tree property if PFA holds). This begs the following questions:

- Are the principles $(\kappa,\lambda)$-STP and $(\kappa,\lambda)$-SSTP equivalent for all $\lambda$ (or some fixed $\lambda$ depending on $\kappa$)?
- Is the super tree property equivalent to the simultaneous super tree property?