The symmetric Stone model has no componentwise proper selector

Suppose $f\in N$ is such a selector. Choose a hereditarily symmetric name $\dot{f}$, a condition $p_0$ forcing that $\dot{f}$ has the stated selector property, and a support $e\subseteq \lambda \times \mathbb{R}\times \lambda$ of size below $\lambda$ such that every member of $\mathrm{fix}(e)$ fixes $\dot{f}$.

The projection of $e$ to the first coordinate has size below $\lambda$, so choose $\xi <\lambda$ outside it. Strengthen $p_0$ to a condition $p$ and choose distinct ground-model reals $r,s$ so that $p$ forces $X_{\xi r}\in \dot{f}(R_{\xi })$ and $X_{\xi s}\mathrm{\notin }\dot{f}(R_{\xi })$.

Let $C$ be the set of third coordinates $\alpha$ occurring in $\mathrm{dom}(p)$ at first coordinate $\xi$. Then $\mathrm{\mid }C\mathrm{\mid }<\lambda$. Choose a disjoint $C^{\mathrm{\prime }}\subseteq \lambda$ of the same cardinality and a permutation $\tau$ of $\lambda$ interchanging $C$ and $C^{\mathrm{\prime }}$ and fixing the complement. Let $\rho$ be reflection about $(r+s)\mathrm{/}2$, and let $\pi$ be the coordinate-local automorphism from [F2] using $\rho$ and $\tau$ at $\xi$ and the identity elsewhere.

The automorphism $\pi$ lies in $\mathrm{fix}(e)$ because it is the identity outside the $\xi$-block, so $\pi \dot{f}=\dot{f}$. It fixes $R_{\xi }$ and swaps $X_{\xi r}$ with $X_{\xi s}$. By [F3], $\pi p$ therefore forces $X_{\xi s}\in \dot{f}(R_{\xi })$ and $X_{\xi r}\mathrm{\notin }\dot{f}(R_{\xi })$.

The conditions $p$ and $\pi p$ are compatible. On every coordinate whose first index is not $\xi$, $\pi$ is the identity, so the two conditions agree on their common domain. At first index $\xi$, every third coordinate used by $p$ lies in $C$, whereas every third coordinate used by $\pi p$ lies in the disjoint set $C^{\mathrm{\prime }}$, so their domains are disjoint there. Hence $p\cup \pi p$ is a common extension.

The common extension $p\cup \pi p$ inherits from $p$ the assertion $X_{\xi s}\mathrm{\notin }\dot{f}(R_{\xi })$ and from $\pi p$ the assertion $X_{\xi s}\in \dot{f}(R_{\xi })$, a contradiction. Therefore no componentwise nonempty proper selector belongs to $N$.