Let $X$ be a spectrum, and let $x : S^0 \to X$ be a map. Let the stable James construction $J(X,x)$ denote the free $E_1$ ring on the $E_0$-ring $(X,x)$. It is computed as the colimit of the $\Delta^{inj}$-indexed diagram $X \rightrightarrows X \otimes X \cdots$ where the coface map $d^i_n : X^{\otimes n-1} \to X^{\otimes n}$ is given by $id_X \otimes \cdots \otimes x \otimes \cdots \otimes id_X$ with $x$ in the $i$th slot. Let $J_n(X)$ denote the colimit of the restriction of this diagram to a $\Delta^{inj}_{\leq n}$-indexed diagram, so that $J(X)$ is the filtered colimit of the $J_n(X)$'s.
Question: Let $Y$ be a finite spectrum. Is it the case that $J(X) \otimes Y = 0$ iff $x^{\otimes n} \otimes id_Y : Y \to X^{\otimes n} \otimes Y$ is null for some $n \in \mathbb N$?
Notes:
Certainly we have that $J(X) \otimes Y = 0$ iff $Y \to J_n(X) \otimes Y$ is null for some $n \in \mathbb N$. If this holds with $n = 1$, then trivially we have $x \otimes Y = 0$ as desired.
If it holds with $n = 2$, then since $J_2(X)$ is the cofiber of $X \xrightarrow{d_2^0 - d_2^1} X \otimes X$, we have a factorization $x \otimes x \otimes id_Y = ((d_2^0 - d_2^1) \otimes id_Y) \circ f$ for some $f : Y \to X \otimes Y$. Then
$$ x^{\otimes 3} = \sum_{i = 0}^2 (-1)^i x^{\otimes 3} = \sum_{i = 0}^2 (-1)^i d_3^i \circ (x \otimes x)$$
$$ x^{\otimes 3} \otimes id_Y = \sum_{i = 0}^2 (-1)^i (d_3^i \otimes id_Y) \circ (x \otimes x \otimes id_Y) $$ $$ = \sum_{i = 0}^2 (-1)^i (d_3^i \otimes id_Y) \circ ((d_2^0 - d_2^1) \otimes id_Y) \circ f = 0$$
The last equation is because we have $(x \otimes x \otimes id_X - x \otimes id_X \otimes x) - (x \otimes x \otimes id_X - id_X \otimes x \otimes x) + (x \otimes id_X \otimes x - id_X \otimes x \otimes x) = 0$.
I suspect this computation generalizes, but as $n$ gets bigger, the expression for $J_n(X)$ in terms of the $X^{\otimes i}$ gets more complicated and I'm not sure how to proceed.
