Suppose $X$ is a finite $n$-connected CW complex, then the function spectrum $F(\Sigma^\infty_+ X,S^0)$ is an $E_\infty$-ring spectrum, induced by the diagonal of $X$. I believe it is known that if we consider $F(\Sigma^\infty_+ X,S^0)$ as an $E_n$-ring spectrum, its Koszul dual is the $E_n$-ring spectrum $\Sigma^\infty_+ \Omega^n X$. Let me sketch an argument:
The Koszul dual $E_n$-ring spectrum is computed as the Spanier-Whitehead dual of the factorization homology $\int_{(\mathbb{R}^n)^+}(-)$. We know factorization homology commutes with taking suspension spectra, so $\int_{(\mathbb{R}^n)^+} \Sigma^\infty_+ \Omega^n X \simeq \Sigma^\infty_+ \int_{(\mathbb{R}^n)^+} \Omega^n X$. By work of Ayala-Francis, the latter is known to be equivalent to $\Sigma^\infty_+ B^n\Omega^n X\simeq \Sigma^\infty_+ X$.
The tricky part is then to figure out the $E_n$-coalgebra structure on the result. The $E_n$-coalgebra structure is obtained from the functoriality of factorization homology with respect to pinch maps. Hence, the $E_n$-coalgebra structure can be computed before taking suspension spectra. From this question, I learned that in a cartesian category like $(\mathrm{Top}_*,\times)$, all $E_n$-coalgebras come from the diagonal, so the $E_n$-coalgebra structure of $X$ induced from factorization homology has to be coming from the diagonal. The dual of this is then the original $E_n$-algebra structure on $F(\Sigma^\infty_+ X,S^0)$.
Is there a usual reference for this fact?
Is the supplied proof correct?