If you add some additional structure of remembering that the domain of these transformations is an abelian group, then there is actually a completely canonical

If you add some additional structure of remembering that the domain of these transformations is an abelian group, then there is actually a completely canonical version of the Fourier transform, whose domain is the dual group - the only non-canonical part is picking labels for elements of the dual group, but there is no other choice of basis functions. (Wikipedia gets a little into this, you can find it under "Pontryagin duality" on the Fourier transform page, but it's not really enough of an explanation to learn from). Take the discrete Fourier transform for example, whose domain group is Z/nZ: the integers-mod-n for some fixed n. The basis functions for the transforms need to be group homomorphisms f: Z/nZ -> C, meaning that they need to satisfy f(a + b) = f(a) f(b) for all a, b. Without too much trouble you can work out that the only such functions are of the form f(k) = exp(2 pi i m k / n) for m = 0, ..., n - 1. So once the group structure is remembered, the choice of basis is forced. The only ambiguity is relabelling the basis elements back into Z/nZ, which is a (very useful) hack. The same reasoning (with more complicated working) applies to taking the Fourier series of a periodic function, or taking the Fourier transform of a continuous function on the real line. There is a canonical choice of basis, but some ambiguity in what labels to assign to those basis elements.