Decision problem for a bundle in a line bundle extension of $\dim Ext^1=1$. Let $E$ be a vector bundle over a nonsingular irreducible surface $X$ given by an extension
$$0 \rightarrow A \rightarrow E \rightarrow B \rightarrow 0,$$
where both $A$ and $B$ are a line bundle over $X$. Then I have known that the extension is classified by the group $Ext^1(B,A)\cong H^1(X,A\otimes B^{*})$ and the split extension $A\oplus B$ is correspondent to the zero element of $H^1(X,A\otimes B^{*})$.
Now suppose $\dim H^1(X,A\otimes B^{*})=1$. Then my question is :
If I had a vector bundle satisfying the same extension to $E$, say $E'$, constructed by some way and proven that it correspondent to $1\in H^1(X,A\otimes B^{*})$ but $E'\ncong E$, then can I conclude that : so $E$ is correspondent to $0\in H^1(X,A\otimes B^{*})$ and thus $E\cong A\oplus B$?
 A: Not unless you're working over $\Bbb F_2$. In any abelian category, there's a bijection between isomorphism classes of extensions of $B$ by $A$ and elements of $\operatorname{Ext}^1(A,B)$ where the split extension corresponds to $0$. Knowing that $E'$ is not isomorphic as an extension to the trivial extension and that $E$ is not  isomorphic to $E'$ as an extension won't force $E$ to be the trivial extension unless $|\operatorname{Ext}^1(A,B)|=2$.
A: The answer to your question is yes.
$E$ and $E’$ arising from extensions of $B$ by $A$ can be isomorphic as $\mathcal{O}_X$-modules, but not isomorphic as extensions of $B$ by $A$, meaning there is $\textit{no}$ isomorphism $f:E\to E’$ making the following diagram commute.
$\require{AMScd}$
\begin{CD}
0@>>>A @>>> E@>>>B@>>>0\\@.@VV{=}V @VVfV@VV{=}V@.\\
0@>>> A @>>>E’@>>>B@>>>0
\end{CD}
So I disagree with Hank’s answer.
Actually, the automorphisms $\text{Aut}(A)\times \text{Aut}(B)$ acts on the $\text{Ext}^1(B,A)$. In each orbit, the isomorphism class of $E$ is unchanged.
In particular, when both $A$ and $B$ are line bundles, $\text{Aut}(A)= \text{Aut}(B)=k^*$, so $k^*\times k^*$ acts on $\text{Ext}^1(B,A)$, with the diagonal acts trivially. So the isomorphism classes of $E$ are parameterized by $\text{Ext}^1(B,A)/k^*$. When $\dim \text{Ext}^1(B,A)=1$, there are just two orbits, with $0$ corresponding to the trivial extension and $1$ corresponding to the nontrivial extension.
A good reference for this is Friedman’s Algebraic Surfaces and Holomorphic Vector Bundles, p.31.
