Ext preserves the tensor

Question

Let $A,B,C$ be three locally free sheaves on some complex manifold or variety $(X,\mathscr O_X)$.

Q: Can we say $Ext^1(A,B)=Ext^1(A\otimes C,B\otimes C)$?

Answer 1

As said in the comments, we need $C$ of rank 1. Otherwise $C=\mathcal{O}_X^{\oplus 2}$ is a counter-example whenever $\operatorname{Ext}^1(A,B)\neq 0$.

But if $C$ is of rank 1, then it is invertible. In particular, the functors $B\mapsto\operatorname{Hom}(A,B)$ and $B\mapsto\operatorname{Hom}(A\otimes C,B\otimes C)$ are naturally isomorphic. Thus, so are their derived functors.