Is a Fourier transform a change of basis, or is it a linear transformation?

I've frequently heard that a Fourier transform is "just a change of basis".

However, I'm not sure whether that's correct, in terms of the terminology of "change of basis" versus "transformation" in linear algebra.

Is a Fourier transform of a function merely a change of basis (i.e. the identity transformation from the original vector space onto itself, with merely a change of basis vectors), or is indeed a linear transformation from the original vector space onto another vector space (the "Fourier" vector space, so to speak)?

Also: Does the same answer hold for other similar transforms (e.g. Laplace transforms)? Or is there a different terminology for those?

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Calling the Fourier transformation "a change of basis" is misleading in the sense that the Fourier transformation is a unitary (linear) transformation between two different Hilbert spaces, namely $L^2(\mathbb R)$ and $L^2(\hat{\mathbb R})$.
Here $\hat{\mathbb R}$ is the dual group of $\mathbb R$. It turns out that $\hat{\mathbb R}\cong\mathbb R$, but there is no canonical isomorphism. So, only if you fix some arbitrary isomorphism $\hat{\mathbb R}\cong\mathbb R$, you can consider the Fourier transformation as a unitary transformation from some Hilbert space to itself, which really is essentially a change of basis.
When is a morphism canonical? The construction of $\hat{\mathbb{R}}$ relies on $\text{Hom}(\mathbb{R},T)$, right? Does this miss any of the structure of $\mathbb{R}$? Don't both not even have a metric which can be matched etc.? – NikolajK Nov 3 '12 at 1:26