(**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 $Aut(N)$.



Can a module be an extension in two really different ways?
It is worth noting some very interesting cases when the answer is yes. An amazing [result by Miyata](https://projecteuclid.org/journals/kyoto-journal-of-mathematics/volume-7/issue-1/Note-on-direct-summands-of-modules/10.1215/kjm/1250524308.full) states that if $R$ is Noetherian and commutative, $M,N$ are finitely generated and $E \cong M\oplus N$, any exact sequence
$ 0 \to M \to E \to N \to 0$ must split!


This holds true slightly more generally, when $R$ is (not necessarity commutative) module-finite over a Noetherian commutative ring. Also, the statement holds for finitely generated pro-finite groups, see Goldstein-Guralnick, J. Group Theory 9 (2006), 317–322.


Added: in fact, this [paper by Janet Striuli](http://www.math.unl.edu/%7Ejstriuli2/research/module2.pdf) may be useful for you. She addressed the question: if two elements $\alpha, \beta \in \text{Ext}^1(M,N)$ give isomorphic extension modules, how close must $\alpha, \beta$ be? Her Theorem 1.2 extend Miyata's result (let $I=0$).


