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.

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 $\partial_*\Sigma^\infty\Delta\colon \partial_* \Sigma^\infty\to \partial_* \Sigma^\infty \mbox{Sq}$. Here Sq denotes the functor Sq$(X)=X\wedge X$. The question is what is $\partial_*\Sigma^\infty\Delta$, 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.

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 $$\mbox{Spectra}(M(1), N(1))\to \mbox{Spectra}(M(1)\wedge O(2), N(2))^{h\Sigma_2} \leftarrow \mbox{Spectra}(M(2), N(2))^{h\Sigma_2}.$$ 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.

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.