A linear representation induces an action on projective space with the same line stabilizers

Since $V$ is finite-dimensional, [F1] identifies the $T$-points of $\mathbb{P}(V^{\vee })$ with isomorphism classes of surjections $V_T^{\vee }\to Q$ with $Q$ invertible. Such a surjection splits locally: on an open where $Q\cong {\mathcal{O}}_T$, a lift of $1$ gives a splitting. After shrinking further, one coefficient of the quotient map is a unit, so elementary changes of basis identify its kernel with ${\mathcal{O}}_T^{\dim V-1}$. Dualizing this local splitting gives a rank-one locally direct summand $Q^{\vee }\hookrightarrow V_T$ by [F3]. Conversely, dualizing a rank-one locally direct summand $L\hookrightarrow V_T$ gives a surjection $V_T^{\vee }\to L^{\vee }$; reflexivity makes these constructions inverse. For $V=0$ there are no such quotients or subbundles over a nonempty $T$.

A line $L\subseteq V$ is a rank-one direct summand: extend a nonzero vector of $L$ to a basis of the finite-dimensional $V$. Its dual $V^{\vee }\twoheadrightarrow L^{\vee }$ is therefore a rank-one quotient, and base change to any $T$ gives the point $[L{]}_T$ represented by $V_T^{\vee }\twoheadrightarrow L_T^{\vee }$.

The affine-local automorphisms supplied by [F2] agree on overlaps by naturality, giving $r(g)$ on $V_T$ for every $g\in G(T)$. On $\mathbb{P}(V)$ define $g\cdot [q]=[q\circ r(g{)}^{-1}]$. This respects quotient isomorphisms and base change, and $(gh)\cdot [q]=[q\circ r(h{)}^{-1}\circ r(g{)}^{-1}]=g\cdot (h\cdot [q])$; the identity acts trivially. By [F1] and [F4] this is a scheme action. Apply the same construction to the dual representation $r^{\vee }(g)=(r(g{)}^{-1}{)}^{\vee }$, which satisfies $r^{\vee }(gh)=r^{\vee }(g)r^{\vee }(h)$, to obtain the action on $\mathbb{P}(V^{\vee })$.

On the line-submodule description of step 1.1 the action has the stated form $L\mapsto r(g)L$: if the quotient $q:V_T^{\vee }\to Q$ corresponds to $L=\mathrm{im}(q^{\vee })$ under the double-dual identification, then $r^{\vee }(g{)}^{-1}=r(g{)}^{\vee }$ and the translate $q\circ r(g{)}^{\vee }$ has dual $r(g)\circ q^{\vee }$ by [F3], whose image is $r(g)L$.

Fix a line $L\subseteq V$ and a $k$-algebra $R$ with an element $g\in G(R)$. By steps 1.1 and 2.2 the point $g\cdot [L{]}_R$ is the class of the quotient $q_L\circ r^{\vee }(g{)}^{-1}=q_L\circ r(g{)}^{\vee }:V_R^{\vee }\to L_R^{\vee }$, and two rank-one locally free quotients of $V_R^{\vee }$ are isomorphic exactly when their kernels agree. The kernel of the quotient attached in step 1.1 to a rank-one subbundle $L^{\mathrm{\prime }}\subseteq V_R$ is its annihilator $L^{\mathrm{\prime }\perp }\subseteq V_R^{\vee }$, so the kernel of $q_L\circ r(g{)}^{\vee }$ is $\{f:f(r(g)v)=0\text{ for all }v\in L\}=(r(g)L_R{)}^{\perp }$. Hence $g$ fixes $[L{]}_R$ exactly when $(r(g)L_R{)}^{\perp }=L_R^{\perp }$, and passing to annihilators in the reflexive finite locally free module $V_R$ of [F3] this is equivalent to $r(g)L_R=L_R$. Therefore the scheme-theoretic stabilizer $G_{[L]}$ has $G_{[L]}(R)=\{g\in G(R):r(g)L_R=L_R\}$ for every $k$-algebra $R$.

If $H\subseteq G$ is a closed subgroup scheme whose functor of points is $H(R)=\{g:r(g)L_R=L_R\}$ for every $R$, then $H$ and $G_{[L]}$ have the same functor of points by step 4.1, so they are equal as closed subschemes of $G$ by the Yoneda lemma [F4]. No smoothness of $G$ or of $H$ was used anywhere in the argument, which completes the proof.