(**Edit:** I've realized that there was an error in my reasoning when I was convincing myself that these two formulations are equivalent. Hailong has given a beautiful affirmative answer to my first question in the case of finite type modules over a noetherian commutative ring. Mariano has given a slick negative answer to the question for non-finite-type modules. Greg has given a beautiful negative answer to my "alternative formulation" even in the finite type case over a noetherian commutative ring. I'm accepting Hailong's answer since that's the one I imagine people will be most immediately interested in if they find this question in the future.)

Suppose we're working the category of modules over some ring $R$. Suppose a module $E$ is an extension of $M$ by $N$ in two different ways. In other words, I have two short exact sequences

\begin{array}{ccccccccc}
 0&\to &N&\xrightarrow{i\_1}&E&\xrightarrow{p\_1}&M&\to &0\\
 & & \wr\downarrow ?& & \wr\downarrow ?& & \wr\downarrow ?\\
 0&\to &N&\xrightarrow{i\_2}&E&\xrightarrow{p\_2}&M&\to &0
\end{array}

> 
> Must there be an isomorphism between these two short exact sequences?
> 
> 
> 

---

Alternative formulation
-----------------------

$Ext^1(M,N)$ parameterizes extensions of $M$ by $N$ modulo isomorphims *of extensions*. Suppose I'm interested in parameterizing extensions of $M$ by $N$ modulo *abstract isomorphisms* (which don't have to respect the submodule $N$ or the quotient $M$). One obvious thing to note is that there is a left action of $Aut(M)$ on $Ext^1(M,N)$, and that any two extensions related by this action are abstractly isomorphic. Similarly, there is a right action of $Aut(N)$ so that any two extensions related by the action are abstractly isomorphic.

> 
> Does the quotient set $Aut(M)\backslash Ext^1(M,N)/Aut(N)$ parameterize extensions of $M$ by $N$ modulo abstract isomorphism?
> 
> 
> 

Note: I'm **not** asking whether all abstract isomorphisms are generated by $Aut(M)$ and $Aut(N)$. They certainly aren't. I'm asking whether for every pair of abstractly isomorphic extensions *there exists* some isomorphism between them which is generated by $Aut(M)$ and $Aut(N)$.