Here is a collection of facts that all seem true, but together seem to give a nonsensical solution:

1. After rationalization, all natural transformations $F \sim G$ between homogenous functors $F,G:Sp \rightarrow Sp$ of different degrees appear trivial, in the sense that $\operatorname{cofiber}(\sim) \simeq G \vee\Sigma F$. This follows from the fact that rationally the Taylor tower splits (for functors from spectra to spectra).

2. If one has compositions of functors $f \circ g, f' \circ g'$ and natural transformations $\alpha: f \rightarrow f',\beta: g \rightarrow g'$ such that either $\alpha$ or $\beta$ is trivial on derivatives, then $\alpha \circ \beta$ is trivial on derivatives. This follows from the chain rule of Arone and Ching that the derivatives can be computed as a (derived) composition product.

3. The natural transformation $\Sigma^\infty \Omega^\infty \rightarrow \Sigma^\infty (\Omega^\infty(-))^{\wedge 2}$ given by $1 \circ \Delta \circ 1$ induced by the diagonal $\Delta$ in $Top_*$ is $0$ on derivatives. This is because $\Sigma^\infty \Delta: \Sigma^\infty \rightarrow \Sigma^\infty(-)^{\wedge 2}$ is a natural transformation between homogenous functors of different degrees.

4. The Taylor tower $(\Sigma^\infty \Omega^\infty )^{\wedge n}$ converges on $0$-connected spectra. I believe this follows from the fact for $n=1$.

5. Hence, for a connected infinite loop space $X$, the cofiber of $\Sigma ^\infty \Delta: \Sigma^ \infty X \rightarrow \Sigma^\infty (X \wedge X)$ is rationally equivalent to $\Sigma^\infty (X \wedge X) \vee \Sigma^\infty X$.

However, if we let $X=K(\mathbb{Q},2k)$, then $X$ has a nontrivial cohomology ring, which implies that $\operatorname{cofiber}(\Sigma^ \infty X \rightarrow \Sigma^\infty (X \wedge X))$ should not be equivalent to $\Sigma^\infty (X \wedge X) \vee \Sigma^\infty X$.

I am wondering in which step there is an error.