Action of group scheme $G$ on vector bundle $\mathbb V(M)$ compatible with scaling. Is it automatically linear?

Fix a commutative ring $k$. $\def\Spec{\operatorname{Spec}}\def\Mod{\operatorname{Mod}}\def\CAlg{\operatorname{CAlg}}\def\Ab{\operatorname{Ab}}\def\Sym{\mathcal{S}}\def\V{\mathbb{V}}\def\A{\mathbb{A}}$ Let $M$ be a $k$-module and $G=\Spec H$ be an affine group scheme. After being initially skeptical, I learned in my last post that there is a well-known equivalence between

• Representations of $G$ on $M$, which are natural transformations $r\colon G\times\tilde M\to\tilde M$, where $\tilde M$ is the functor $\CAlg_k\to\Ab$ given by $R\mapsto M\otimes_kR$ and $G$ is seen as the functor $R\mapsto\hom_{\CAlg_k}(H,R)$, such that each component $G(R)\times(M\otimes_kR)\to(M\otimes_kR)$ is an $R$-linear action.
• $H$-comodule structures on $M$, which are $k$-linear maps $\rho\colon M\to M\otimes_kH$ satisfying coassociativity and counitality (with respect to the Hopf algebra structure of $H$).

I want to compare these two notions with a third. Besides $\tilde M\colon R\mapsto M\otimes_kR$, there is another easy functor $\CAlg_k\to\Ab$ associated with $M$, namely $\mathbb V(M)\colon R\mapsto \hom_{\Mod_k}(M,R)$. In fact, since $$\hom_{\Mod_k}(M,R)=\hom_{\CAlg_k}(\Sym^* M,R)$$ it turns out that $\mathbb V(M)$ is the affine $k$-scheme $\mathbb V(M)=\Spec(\Sym^*M)$. This scheme has a natural scalar multiplication map $\A^{1}\times\V(M)\to\V(M)$ and an addition map $\V(M)\times\V(M)\to\V(M)$. The group scheme $G$ can act on any $k$-scheme $Y$ via a morphism of schemes $G\times Y\to Y$, so we can define a "geometric" linear action of $G$ as an action of $G$ on $\V(M)$ which is compatible with scaling and addition.

In this MSE answer it is stated (without elaboration) that this third notion is equivalent (at least in the case that $k$ is a field) to the other two, so I tried to prove it.

My attempt

Since $\V(M)=\Spec\Sym^*(M)$, the action $G\times\V(M)\to\V(M)$ is equivalent to a coaction of $H$ on the algebra $\Sym^*M$: $$\Sym^*M\to H\otimes_k\Sym^*M$$ Being a map of algebras, the universal property of the symmetric algebra says that such a coaction is also equivalent to a $k$-linear map $$\rho\colon M\to H\otimes_k\Sym^*M.$$ Now, my idea was to prove that the action $G\times\V(M)\to\V(M)$ is linear if, and only if, the image of $\rho$ is contained in $H\otimes_kM$, giving rise to a coaction $M\to H\otimes_k M$. For this, I need to know what condition on $\rho$ is imposed by requiring linearity of the action.

The scaling action $\V(M)\times\A^{1}\to\V(M)$ (I write it on the right avoid a braiding with $G$) and the addition map $\V(M)\times\V(M)\to\V(M)$ also correspond to maps of algebras $$\Sym^*M\to \Sym^*M\otimes k[X]$$ $$\Sym^*M\to \Sym^*M\otimes \Sym^*M$$ which in turn correspond to $k$-linear maps $$\sigma\colon M\to \Sym^*M\otimes k[X]$$ $$\alpha\colon M\to \Sym^*M\otimes \Sym^*M$$ that, according to my calculations, are given by: $$\alpha(m)=m\otimes X,$$ $$\sigma(m)=m\otimes 1 + 1\otimes m.$$ Now write $\rho(m)=\sum x_i \otimes m_{i,1}\cdots m_{i,n_i}$. When I imposed the compatibility of $\rho$ with $\alpha$ and $\sigma$, (in the appropiate sense, mimicking linearity) I got the following equations: \begin{equation} \left\{ \begin{aligned} &\sum_i x_i \otimes m_{i,1}\cdots m_{i,n_i} \otimes X^{n_i} = \sum_i x_i \otimes m_{i,1}\cdots m_{i,n_i}\otimes X \\ &\sum_i x_i \otimes (m_{i,1}\otimes 1 + 1\otimes m_{i,1})\cdots (m_{i,n_i}\otimes 1 + 1\otimes m_{i,n_i}) \\ \qquad &= \sum_i x_i \otimes m_{i,1}\cdots m_{i,n_i}\otimes 1 + x_i \otimes 1\otimes m_{i,1}\cdots m_{i,n_i} \\ \end{aligned} \right. \end{equation} Now comes my issue. Note that the first condition is satisfied iff $n_i=1$ for all $i$, i.e. iff $\rho$ has the form $\rho(m) = \sum_i x_i \otimes m_{i}$. This reduces the second equation to \begin{equation} \sum_i x_i \otimes (m_{i}\otimes 1 + 1\otimes m_{i}) = \sum_i x_i \otimes m_{i}\otimes 1 + x_i \otimes 1\otimes m_{i} \end{equation} but this is always true! so, this would seem to indicate that, if an action is compatible with scaling, then it is also automatically compatible with addition. I wasn't able to find a mistake in my calculations (so far), so I came here to ask: is this true?

If an action of a group scheme $G$ on $\V(M)$ is compatible with scaling, is it also automatically compatible with addition?

It sounds kind of weird, since in basic linear algebra one learns that are functions satisfying $f(av)=af(v)$ but not $f(v+w)=f(v)+f(w)$, however when I tried to port such an example to the current situation, I realized that these examples from linear algebra are all using absolute values, floor functions, square roots or some non-polynomial stuff. Since everything in sight is a scheme and the action is a morphism of schemes, I thought it may be possible that the rigidity of ``algebraic'' maps was the culprit of this weird result.

Of course (and I realise this may be too much to ask), if I made a mistake, I would like to know where it is!

Thanks.