$ \mathrm{ch}(F(A)) = \mathrm{ch}(F(B)) \implies \mathrm{ch}(A) = \mathrm{ch}(B) $ for autoequivalences?

Question

Let $\mathcal{C} \subset D^b(X)$ be a subcategory of the derived category of coherent sheaves on a smooth projective variety $X$. Let $F : \mathcal{C} \to \mathcal{C}$ be an autoequivalence. Let $\mathrm{ch}(-)$ denote the Chern character of an object in $D^b(X)$. Let $A, B \in \mathcal{C}$ be objects. Do we have the implication $$ \mathrm{ch}(F(A)) = \mathrm{ch}(F(B)) \implies \mathrm{ch}(A) = \mathrm{ch}(B) \, \, ? $$ For easy autoequivalences such as shifting or twisting by e.g. $\mathcal{O}_X(H)$ (I guess these are autoequivalences of the whole category $D^b(X)$, i.e. $\mathcal{C} = D^b(X)$), the implication holds but I'm not sure how to show whether it's true for an arbitrary autoequivalence/what conditions to put on $F$?