Tannakian reconstruction and the distribution algebra

Question

$\DeclareMathOperator\Dist{Dist}\DeclareMathOperator\Lie{Lie}\DeclareMathOperator\Rep{Rep}\DeclareMathOperator\End{End}$Let $G$ be an affine group scheme over a commutative ring $k$ (I am mainly interested in the case where $G$ is split reductive, say $G=\operatorname{GL}_n$, but I want to allow $k$ to not be field, say $k=\mathbb{Z}$).

Let $\mathcal{O}(G)$ be the coordinate ring of $G$ with augmentation ideal $\mathfrak{m}$ and coproduct $\Delta$. The distribution algebra $\Dist(G)$ is the $k$-module of $k$-linear maps $\mathcal{O}(G)\to k$ that vanish on $\mathfrak{m}^n$ for some $n$, with product $\mu \cdot \lambda = (\mu \otimes \lambda) \circ \Delta$ (when $k$ is a characteristic zero field, this is the same as the universal enveloping algebra of $\Lie(G)$). Let $\Dist^{\le n}(G)$ be the distributions which vanish on $\mathfrak{m}^{n+1}$, which make $\Dist(G)$ into a filtered algebra.

In the spirit of Tannakian reconstruction, I want to know how one can recover $\Dist(G)$ directly from the Tannakian category $\Rep_G$ of representations of $G$ on finitely generated $k$-modules with its forgetful functor $F$ to $k$-modules. Every representation of $G$ has a naturally induced action of $\Dist(G)$ (see Jantzen's Representations of algebraic groups, I.7), so there is a map $\Dist(G) \to \End(F)$, which I believe is always injective (at least in nice cases).

How can we characterize the image of $\Dist(G)$ inside $\End(F)$?

For example, if $G=\mathbb{G}_{m}$, then $\Rep_{G}$ is the category of $\mathbb{Z}$-graded vectors spaces, and an endomorphism of $F$ is the same as a $\mathbb{Z}$-indexed sequence $(a_n)_{n\in \mathbb{Z}}\in k^\mathbb{Z}$, where the endomorphism acts on the degree $n$ part of a graded vector space via multiplication by $a_n$. The endomorphisms coming from $\Dist(G)$ are exactly those for which $a_n$ is a polynomial function of $n$ (where by polynomial function I mean that applying the discrete difference operator enough times gives zero, though it might not be given by a polynomial with coefficients in $k$, e.g. if $k=\mathbb{Z}$ then $a_n=\frac{n(n-1)}{2}$ works).

Here is a guess of what the answer might be, though I'm not sure how to prove it if it's true. I know that the elements of $\Lie(G)$ correspond to natural endomorphisms $(\phi_V)_{V \in \Rep_G}$ satisfying $$\phi_{V_1\otimes V_2} = 1_{V_1}\otimes \phi_{V_2} + \phi_{V_1} \otimes 1_{V_2}$$ (Tannakian theory for Lie algebras). More generally, the elements of $\Dist^{\le 1}(G)=k \oplus \Lie(G)$ correspond to natural endomorphisms satisfying $$\phi_{V_1\otimes V_2} - 1_{V_1}\otimes \phi_{V_2} - \phi_{V_1} \otimes 1_{V_2} + \phi_k 1_{V_1}\otimes 1_{V_2}=0.$$ This suggests that the elements of $\Dist^{\le n}(G)$ might correspond exactly to the natural endomorphisms satisfying $$\sum_{S\subset [n]} (-1)^{\lvert S\rvert} \phi_{V_1, \cdots, V_n}^S=0$$ where, for $S=\{i_1 < \dotsb< i_r\} \subset \{1,\dotsc, n\}$, $\phi_{V_1, \dotsc, V_n}^S$ is the endomorphism of $V_1 \otimes \dotsb \otimes V_n$ which acts by $\phi_{V_{i_1} \otimes \dotsb \otimes V_{i_r}}$ on $V_{i_1} \otimes \dotsb \otimes V_{i_r}$ and by the identity on all the other factors. This does give the right answer in the $\mathbb{G}_m$ example above.