Let $s \geq 0$ be fixed. The $J$-homomorphism includes $\pi_{8s+1}(SO) = \mathbb Z/2$ in $\pi_{8s+1}^s$, the $(8s+1)$-th stable homotopy group of spheres.

Now regard $\pi_{8s+1}^s = \pi_{8s+1} ((B\Sigma_{\infty})^+)$ , where $\Sigma_{\infty}$ is the group of compactly supported permutations of $\mathbb N$.

The obvious map (i.e. write permutations as permutation matrices) $\Sigma_{\infty} \to O$ induces a map $\pi_{8s+1}^s \to \pi_{8s+1}(BO) = \pi_{8s}(O)= \mathbb Z/2$.

Is the composition $\mathbb Z/2 = \pi_{8s+1} (SO) \to \pi_{8s+1}^s \to \pi_{8s} (O) = \mathbb Z/2$ trivial or nontrivial?

For $s = 0$ this map is nontrivial, since the two elements of $\pi_1^s$ correspond to odd and even permutations in $\Sigma_{\infty}$, whereas the two elements in $\pi_1(BO) = \pi_0(O)$ correspond to determinant $\pm1$ matrices and the map $\Sigma_{\infty} \to O$ preserves this structure.

So what happens for positive $s$? [I have a strong feeling that the map is then trivial, but maybe I am wrong...]

Remark: This map factors through $K_{8s+1}(\mathbb Z)$.