This question is about Section 5.d in Milne's notes on algebraic groups (see https://www.jmilne.org/math/CourseNotes/iAG200.pdf), and more specifically on the proof of Lemma 5.25 in the case when $G$ is not smooth.
At some point there is the claim that $L^{\otimes p^n} \subseteq (V^{\otimes p^n})^N$ and I don't understand why this should hold.
To illustrate that there might be a problem, let's consider the following degenerate situation: Let $k$ be of positive characteristic $p$, let $G = \alpha_p$ and let $H \subseteq G$ be the trivial subgroup. Let $V = k^2$ that we turn into a $G$-representation via $$ a \mapsto \begin{pmatrix} 1 & 0 \\ a & 1 \end{pmatrix} $$ and consider the line $L = k \cdot e_1 \subseteq V$ that is clearly stable under $H$. In this setting we can choose $n = 1$ because the image of $F \colon G \to G$ is already trivial (so smooth). The claim is then that $N = \ker(F \colon G \to G) = G$ stabilizes $L^{\otimes p} \subseteq V^{\otimes p}$ and acts trivially on it. But I think this is not true as for example the group element $g = \varepsilon \in G(k[\varepsilon]/(\varepsilon^2))$ acts on $e_1^{\otimes p}$ as $$ g \cdot e_1^{\otimes p} = (e_1 + \varepsilon e_2)^{\otimes p} = e_1^{\otimes p} + \sum_{i = 1}^p \varepsilon \cdot (e_1^{\otimes (i-1)} \otimes e_2 \otimes e_1^{\otimes (p-i)}) \neq e_1^{\otimes p}. $$ Of course in this example $H$ only has the trivial character so that the Lemma is true for trivial reasons, but still the example illustrates that there might be some problem with the proof.
Also I think that this problem can be solved by working with symmetric tensors instead (it certainly fixes my example, but I think the whole proof should work then).