We know stable bundles have a good property:

If $f: E\longrightarrow E'$ is a nontrivial morphism where $E,E'$ have the same rank and degree, then $f$ must be an isomorphism.

I'm wondering does this property hold for simple bundles? If not, are there any counter-examples? i.e, find two simple bundles $E,E'$ of same rank and degree but there exists a non-trivial morphism $f$ between them which is not isomorphic.

This property is the key to the Hausdorff property of the moduli space of stable bundles, say $\mathcal{M}^s_{X}(n,d)$, where $n$ is the rank, $d$ is the degree and $X$ is a genus $g$ curve. Since for stable bundles $E_1,E_0$, if $\{E_t\}$ is a family of isomorphism classes of $E_1$ which deforms to $E_0$, then it defines a nontrivial morphism $E_1\longrightarrow E_0$, hence it must be isomorphism, that eliminates the jumping and non-Hausdorff phenomenon.

But for simple bundles, It was showed in Remark 12.3 of Narasimhan-Seshadri's paper that the moduli space $\mathcal{M}^{\text{sim}}_{X}(2,1)$ of rank 2 degree 1 simple bundles over a genus 3 curve $X$ is not Hausdorff, so I think a counter-example of the above property may occur in this case.