I love it when an undergraduate catches me out. I'm lecturing a first course in (not necessarily commutative) rings (with 1) and I've spent the last few weeks doing basic module theory. I defined a short exact sequence of (left) $R$-modules and a homomorphism of short exact sequences (a homomorphism from $0\to A\to B\to C\to 0$ to $0\to A'\to B'\to C'\to 0$ is just $R$-module maps $A\to A'$, $B\to B'$ and $C\to C'$ such that the obvious two squares commute). There's hence an obvious notion of an isomorphism of short exact sequences.
Today one of the students asked me if it was possible to have a ring $R$ and modules $A,A',B,B',C,C'$ sitting in two short exact sequences as above, and such that $A$ was isomorphic to $A'$, $B$ was isomorphic to $B'$ and $C$ was isomorphic to $C'$, but that the sequences weren't isomorphic. I said "sure, I'll email you a counterexample later" (the logic being that if this were a theorem, it would be one I knew about). I thought I'd knock up a counterexample on the tube home -- but I failed :-/
If $R$ is a field then short exact sequences split (yes we're assuming AC) so that's not going anywhere. So I thought that $R=k[X]$ would be a good place to start, $k$ a field. In this case an $R$-module is just a $k$-vector space equipped with an endomorphism and I figured this would give me enough flexibility. I wanted $A,A',C,C'$ to be $R/(x^2)$ and tried some messy matrix calculations to figure out an example, but I couldn't get it to work. I then went for $R=k[x,y]$ but now a 2-dimensional vector space is an $R$-module when we give it two commuting linear maps and somehow this set-up had too many endomorphisms for me to face doing the algebra. I then figured that I might want to try a polynomial ring in two noncommuting variables--but then it was my stop and it was time to start thinking about other things.
I am almost certain that there will even be a counterexample with $R$ commutative (that's why I was thinking about the commutative case). Can anyone tell me the trick I'm missing?