Shelah proved (paper 667) that if GCH holds and $\lambda$ is singular, then for every stationary $S \subseteq \{ \alpha < \lambda^+ : \text{cf}(\alpha) = \text{cf}(\lambda) \}$, there is a ladder system $\{ c_\alpha : \alpha \in S \}$ and a coloring of the ladders $\{ f_\alpha : \alpha \in S \}$ in 2 colors (where each $f_\alpha : c_\alpha \to 2$), which cannot be uniformized. This means that there is not any $F : \lambda^+ \to 2$ such that for all $\alpha \in S$, there is a set $b_\alpha$ of size $<\text{cf}(\lambda)$ such that $F \restriction (c_\alpha \setminus b_\alpha) = f_\alpha \restriction (c_\alpha \setminus b_\alpha)$. This is in contrast to the case of $\lambda$ regular, where it is always possible to force ladder system uniformization on an appropriately stationary-costationary set and preserve GCH.
Shelah built the ladder roughly by diagonalizing against all small approximations, making sure that for all $F : \lambda^+ \to 2$, there is $\alpha \in S$ such that, if $\langle\beta_i : i < \text{cf}(\lambda) \rangle$ enumerates $c_\alpha$, then $F(\beta_{2i}) = F(\beta_{2i+1})$ for all $i<\text{cf}(\lambda)$. This guarantees that the “even-odd” coloring of the ladders cannot agree mod bounded with a single $F: \lambda^+ \to 2$.
Question: Is it consistent with GCH that there is a singular $\lambda$ and a stationary $S \subseteq \{ \alpha < \lambda^+ : \text{cf}(\alpha) = \text{cf}(\lambda) \}$ such that for all ladder systems $\langle c_\alpha : \alpha \in S \rangle$ and all “constant colorings” $\langle f_\alpha : \alpha \in S \rangle$ (where each $f_\alpha : c_\alpha \to 2$ is constant), there is a uniformization $F : \lambda^+ \to 2$?