TITLE: (Short) Exact sequences with no commutative diagram between them
QUESTION [32 upvotes]: This question was asked by a student (in a slightly different form), and I was unable to answer it properly. I think it's quite interesting.
The problem is to produce an example of the following situation: find a short exact sequence
$$ 0 \to X_1 \to X_2 \to X_3 \to 0$$
(in some category of your choice), and a second exact sequence
$$ 0 \to Y_1 \to Y_2 \to Y_3 \to 0$$
such that there are isomorphisms $X_n \cong Y_n$ for all $n$, BUT in such a way that there is no commutative diagram whatsoever between the two sequences, with the vertical maps being isomorphisms.
It is impossible to find such an example in the category of vector spaces, or of finitely-generated abelian groups. I don't know about the general case, though.
I would be grateful for any example, but would be disappointed if the chosen category were constructed specifically to answer the problem.
EDIT / COMMENT: in the first version of this question I was asking for sequences which could potentially be infinite. Some great examples came in the comments straightaway. I'm very thankful for them, but I recall only now that the student's original question was about short exact sequences, so I've edited accordingly. (I'm sorry for the confusion, the student asked me this question several months ago, and I was posting only now, for some reason... and got it wrong. I'm very happy to know about the examples involving infinite sequences, though.)

REPLY [4 votes]: Here is an answer involving finitely generated groups in the quotient, and free groups in the kernel and the total group. 
Suppose that $G$ is an infinite, finitely generated group. Let $F_n = \langle s_1,...,s_n \rangle$ be the rank $n$ free group, and let $f : F_n \to G$ be a surjective, noninjective homomorphism; this is a fancy way of saying that $f(s_1),...,f(s_n)$ is a generating set of $G$ but not a free basis. The kernel of $f$, being an infinite index, nontrivial, normal subgroup of $F_n$, must be of infinite rank. So, you get a short exact sequence
$$1 \to F_\infty \to F_n \xrightarrow{f} G \to 1
$$
Now we just have to dream up some $G$ and two homomorphisms $f,g : F_n \to G$ for which the two short exact sequences are inequivalent. This problem is a translation of a well-studied (but nonetheless still mysterious) problem in group theory: the two short exact sequences are equivalent if and only if the generating sets $f(s_1),...,f(s_n)$ and $g(s_1),...,g(s_n)$ are ``Nielsen equivalent''. 
This paper (by Lars Louder with most cases attributed to Zieschang) shows that if $G$ is the fundamental group of a closed surface of genus $g \ge 2$ then any generating set of cardinality $2g$ is Nielsen equivalent to the "standard" one given by the presentation
$$\pi_1(G) = \langle a_1,b_1,...,a_g,b_g \, | \,\, [a_1,b_1] [a_2,b_2] ... [a_g,b_g] = 1\rangle
$$
So we won't get an answer from surface groups.
On the other hand the paper of Lustig and Moriah (in Topology (1991)) gives examples where $G$ is a Fuchsian group with torsion (fundamental group of a closed 2-D orbifold) and two generating sets of the same cardinality, both of which look "standard", but which are not Nielsen equivalent. 
This paper (by Kapovich and Weidmann) has an introduction with a survey of examples.