TITLE: Constructions unique up to non-unique isomorphism
QUESTION [24 upvotes]: 1) Fields have algebraic closures unique up to a non-unique isomorphism.
2) Nice spaces (without base point) have universal covering spaces unique up to a non-unique isomorphism.
3) Modules have injective hulls unique up to a non-unique isomorphism.
Such situations can lead to interesting groups - the absolute Galois group, the fundamental group, and the "Galois" groups of modules introduced by Sylvia Wiegand in Can. J. Math., Vol. XXIV, No. 4, 1972, pp. 573-579.
I'd appreciate any insight into the abstract features of situations which give rise to this type of phenomenon.  And I'd appreciate as many examples from as many parts of mathematics as possible.

REPLY [4 votes]: Let me mention Sullivan's minimal models.
Every commutative differential graded $\mathbb{Q}$-algebra (cdga) $A^*$ concentrated in non-negative degrees and such that $H^0(A^*)=\mathbb{Q}$ admits a minimal Sullivan model $i:M^*\to A^*$ where $M^*$ is a free commutative graded algebra obtained from $\mathbb{Q}$ by adding generators of non-negative degrees so that the differential of each generator is a $\mathbb{Q}$-linear combination of products of length $\geq 2$ of the previous generators, and $i$ is a map of cgda's that induces a cohomology isomorphism (i.e., a quasi-isomorphism).
The minimal model is unique up to a non-unique isomorphism. More generally, if $f:A^*\to B^*$ is a map of cdga's and $j:N^*\to B^*$ is a minimal model of $B^*$, then there is a cdga map $g:M^*\to N^*$, defined up to cdga homotopy, such that $fi=gj$ up to cdga homotopy; moreover, if $f$ is a quasi-isomorphism, then $g$ is an isomorphism.
This reduces the classification of non-negative cdga's up to quasi-isomorphism (and as a consequence, the classification of simply connected topological spaces up to rational homotopy) to the classification of algebras of a certain kind up to isomorphism.
Of course, this example is similar to some mentioned before (in a sense it is the commutative analog of the answer of John Palmieri).