No, usually not. The quotient of the "standard action" of $\{\pm 1\}$ on $\mathbb{P}^2$ is a singular quadric cone $Q$ in $\mathbb{P}^3$. The ideal sheaf $F'$ of a line in this cone is a coherent sheaf. The pullback of this coherent sheaf to $\mathbb{P}^2$ has determinant $\mathcal{O}(-1)$. But there is no invertible sheaf on $Q$ whose pullback to $\mathbb{P}^2$ equals $\mathcal{O}(-1)$.