This is an elaboration on my comment above. Let us consider the natural transformation induced by the diagonal $$\Sigma^\infty\Delta\colon \Sigma^\infty X \to \Sigma^\infty X\wedge X.$$ The natural transformation $\Sigma^\infty \Delta$ induces a map between the Goodwillie derivatives of the functors. Let us denote this induced map by $\partial_*\Sigma^\infty\colon \partial_* \Sigma^\infty\to \partial_* \Sigma^\infty \mbox{Sq}$. Here Sq denotes the functor Sq$(X)=X\wedge X$. The question is what is $\partial_*\Sigma^\infty$, and in particular if it can be non-trivial.
The functor $\Sigma^\infty$ is homogeneous linear, and $\Sigma^\infty$Sq is homogeneous quadratic. So we have the following description of the derivatives as symmetric sequences
$$\partial_*\Sigma^\infty = (\Sigma^\infty S^0, *, *, \ldots, )$$ $$\partial_* \Sigma^\infty\mbox{Sq} = (*, \Sigma^\infty{\Sigma_2}_+, *, *, \ldots, ).$$ The space of maps from $\partial_*\Sigma^\infty$ to $\partial_* \Sigma^\infty\mbox{Sq}$ in the $\infty$-category of symmetric sequences is easily seen to be contractible. So it seems that $\partial_*\Sigma^\infty \Delta$ can only be the trivial map. However, we really are interested in the mapping space in the category of right modules over $\partial_*Id$. Let's see what this means. Let $O=\{O(n)\}$ be an operad in spectra. We consider reduced operads, which means that $O(0)=*$ and $O(1)=\Sigma^\infty S^0$. All our modules are truncated at $2$, so only $O(1)$ and $O(2)$ are relevant. Let $M(1), M(2)$ and $N(1), N(2)$ be (truncated) right modules over $O$. This means that we have $\Sigma_2$-equivariant maps $M(1)\wedge O(2) \to M(2)$ and $N(1)\wedge O(2) \to N(2)$. The mapping spectrum of $O$-module maps from $M$ to $N$ is equivalent to the homotopy pullback of the following diagram $\require{AMScd}$ $$\begin{CD} @. \mbox{Spectra}(M(2), N(2))^{h\Sigma_2}\\ @. @VVV\\ \mbox{Spectra}(M(1), N(1)) @>>> \mbox{Spectra}(M(1)\wedge O(2), N(2))^{h\Sigma_2} \end{CD}$$ By contrast, the space of maps from $M$ to $N$ in the category of symmetric sequences is the product $$\mbox{Spectra}(M(1), N(1))\times \mbox{Spectra}(M(2), N(2))^{h\Sigma_2}.$$ There is an obvious forgetful map from the former mapping spectrum to the latter, which sends the middle term to a point.
In the case of interest, $O=\partial_*Id$, so $O(1)=\Sigma^\infty S^0$ and $O(2)\simeq\Sigma^{-1}\Sigma^\infty S^0$, where $\Sigma_2$ acts trivially on $O(2)$. The modules are $(M(1), M(2))=(S^0, *)$ and $(N(1), N(2)) = (*, \Sigma^\infty {\Sigma_2}_+)$. So the spectrum of $O$-module maps from $M$ to $N$ is the following pullback
$$\mbox{Spectra}(\Sigma^\infty S^0, *)\to \mbox{Spectra}(\Sigma^{-1}\Sigma^\infty S^0, \Sigma^\infty {\Sigma_2}_+)^{h\Sigma_2} \leftarrow \mbox{Spectra}(*, \Sigma^\infty{\Sigma_2}_+)^{h\Sigma_2}.$$ This homotopy pullback is easily seen to be equivalent to $\Sigma^\infty S^0$.
It remains to show that the transformation $\Sigma^\infty\Delta$ actually induces a non-trivial map on derivatives. Here is one way to argue this. It is an exercise in Yoneda that the spectrum of natural transformations from $\Sigma^\infty X$ to $\Sigma^\infty X\wedge X$ is equivalent to the sphere spectrum, with the diagonal corresponding to a generator. There is a natural map of mapping spectra $$\mbox{Nat}(\Sigma^\infty X, \Sigma^\infty X\wedge X)\to \partial_*Id\!-\! mod(\partial_*\Sigma^\infty X, \partial_*\Sigma^\infty X\wedge X).$$ Both the source and the target are equivalent to $\Sigma^\infty S^0$. One can show that for polynomial functors whose derivatives are finite free spectra (or more generally, whose derivatives have vanishing Tate homology) this map is an equivalence.