Let $G$ be a reductive group and $X$ a smooth $G$-variety. Then the fixed point subvariety $X^G$ is also smooth (this is theorem 13.1 of Milne's book on algebraic groups). Suppose in addition that the canonical bundle $\omega_X$ is trivial. Then is $\omega_{X^G}$ also trivial?

One idea I had was to use the adjunction formula, which gives that $\omega_{X^G} = \det(I/I^2)^\vee$, where $I$ is the ideal sheaf of $X^G\subset X$. When $X = \text{Spec}(A)$ is affine, we have $I=(g\cdot f - f)_{g\in G, f\in A}$, but in the general case I'm having trouble giving a sufficiently effective description of $I/I^2$ to calculate its determinant.