Let $X$ be the projective variety obtained by gluing the two points $0, \infty \in \mathbb P^1$ transversally, which means that $X$ is isomorphic to $\{y^2z = x^2(x+z)\} \subset \mathbb P^2$, the nodal rational curve. Consider the natural action of $\mathbb G_m$ on $\mathbb P^1$, given by $\mu.[x:y] =[\mu x: y]$. Since $0=[0:1]$ and $\infty = [1:0]$ are both fixed points, this action descends to an action on $X$. In this mathoverflow answer it is explained why no line bundle $L \in \operatorname{Pic}(X)$ can admit a $\mathbb G_m$-linearization.
However to me it seems that this contradicts the following proposition in Mumford's GIT book:
Proposition 1.5. Let a connected linear algebraic group $G$ act on an algebraic variety $X$, proper over $k$. Let $L$ be an invertible sheaf on $X$, and let $\lambda$ be the $k$-rational point of the Picard scheme $\operatorname{Pic}(X/k)$ defined by $L$. Then some power $L^n$ of $L$ is $G$-linearizable if and only if some multiple $n \lambda$ of $\lambda$ is left fixed by $G$.
If we take $L = \omega_X$ to be the dualizing sheaf (which is a line bundle because $X$ is locally a complete intersection), then I think that any automorphism $\varphi: X \to X$ should induce an isomorphism $\varphi^* \omega_X \to \omega_X$ (is this true? it might be a weak point). This means that the point $\lambda = [\omega_X] \in \operatorname{Pic}(X/k)$ is a fixed point for any group action. So by Mumford's proposition (at least some power of) $\omega_X$ admits a $\mathbb G_m$-linearization.
What's wrong with my reasoning here? I have to admit that my knowledge about the Picard scheme is quite limited, so maybe I didn't understand the proposition correctly?
The linked answer proves that there is no ample line bundle on $X$ that admits a $\mathbb G_m$-linearization. The dualizing sheaf on the rational nodal curve has nonpositive degree hence is not ample.