Autoequivalence of $D^b(\text{coh }X)$ taking a rank $n$ vector bundle V to a torsion sheaf.

Let $E$ be an elliptic curve, $V$ an indecomposable coherent vector bundle. I know how to construct an autoequivalence of $D^b(\text{coh }X)$ that takes $V$ to a torsion sheaf if $V$ has rank $1$. Is there an autoequivalence that takes $V$ to some torsion sheaf if $V$ has arbitrary rank?

-

Suppose $V$ is as in the question.
There exists an injection $\mathcal{L}\to V$ with $\mathcal{L}$ locally free coherent of rank $1$. Aplying a composition of autoequivalences of the form $-\otimes\pm x$ for $x\in E$ we can assume $\mathcal{L}=\mathcal{O}(p)$ for some $p\in E$. The spherical twist $L_{\mathcal{O}}$ is an autoequivalence and takes $\mathcal{O}(p)$ to $\mathcal{O}_p$. Denote the composition of these autoequivalences by $F$. Injectivity is preserved by $F$. Thus the image of $V$ under $F$ has a torsion subsheaf. As $V$ was indecomposable and $F^{-1}$ is commutes with direct sums, $F(V)$ is torsion.
EDIT: This is incorrect. The injectivity argument is wrong. I think a correct version would be: if $F$ denotes the autoequivalence $D^b(X)\to D^b(X)$, then $F$ maps semi-stable sheaves to semi-stable sheaves. $F(\mathcal{L})\to F(V)$ is not zero and thus corresponds to a nonzero morphism between semi-stable sheaves. As $F(\mathcal{L})$ is torsion, $F(V)$ is torsion.