I'm using Dolgachev's book on invariant theory to learn linearizations of group actions. Here is a sketch of main construction: let linear algebraic group $G$ act on a quasi-projective variety $X$, and let $L$ be a very ample line bundle that provide a linearization of the action, then $G$ acts on $W=H^0(X, L)$ by
$$
(g \cdot s)(x)=gs(g^{-1}x)
$$
and the map
$$
X \to \mathbb{P}(W^\vee)
$$
that sends $x \in X$ to the hyperplane $\{s \in W | s(x)=0\} \subset W$ is equivariant, where action of $G$ on $W^\vee$ (we think about $W^\vee$ as a space of hyperplanes) is dual to the action on $W$ that is given by
$$
g \cdot H=g^{-1}H,
$$
for a hyperplane $H \in W^\vee$.
I don't understand one of the steps in the proof that this map is equivariant. The following sequences of identities is used $$ \{s \in W | s(g \cdot x)=0\}=\{s \in W | g^{-1}s(g \cdot x)=0\}=\{s \in W | (g^{-1} \cdot s)(x)=0\}=g^{-1}\{s \in W |s(x)=0 \}=g \cdot\{s \in W | s(g \cdot x)=0\} $$ to conclude that the map is equivariant. I don't understand why $$ \{s \in W | (g^{-1} \cdot s)(x)=0\}=g^{-1}\{s \in W |s(x)=0 \}, $$ I think that it should be $$ \{s \in W | (g^{-1} \cdot s)(x)=0\}=g\{s \in W |s(x)=0 \}, $$ but then the map is not equivariant.
I am getting the same answer that you are.
I wonder if the author intended to use $W = H^0(X,L)^{\vee}$ to carry out this construction.