Actually, there is an exact meaning, but it is not always used in that sense. For two functors $\mathsf F,\mathsf G:\mathscr A\to \mathscr B$ a *natural transformation* is a morphism of functors $\eta:\mathsf F\to\mathsf G$ that is compatible with the functors in the obvious (sic!) way.

For instance if $\mathsf F={\rm id}$ is the identity and $\mathsf G=(\underline{\quad} )^{*}$ is the dual of *finite-dimensional* vector spaces, then even though $\mathsf F(V)\simeq \mathsf G(V)$, there is no natural transformation between $\mathsf F$ and $\mathsf G$ that gives this isomorphism. On the other hand $\mathsf F$ is naturally isomorphic to $\mathsf D:=\mathsf G\circ\mathsf G$ via the natural transformation induced by the usual map to the double dual.

Of course, often people say "there is a natural choice of" whatever. That usually means that the "choice" actually does not involve a "choice". In other words, two different people would be expected to make the same choice.

There is however a danger involved that some authors overlook. There are situations when there are more than one natural choices to make. In other words, just because two choices are both natural, one should not assume that they are necessarily the same. For instance, often one can end up with $(-1)$-times the other choice (say for a map) and then whether they are equal or their sum is zero makes a big difference. For an example, consider choosing a generator of the infinite cyclic group, a.k.a., $(\mathbb Z,+)$. One might be led to believe that "the" natural choice of a generator for $(\mathbb Z,+)$ is $1\in \mathbb Z$, but as long as this is only a *group* there is no way to distinguish $1\in \mathbb Z$ and $-1\in \mathbb Z$. They both generate the group and they are each other's inverses. In other words, there are two *natural* choices of a generator of the *group* $(\mathbb Z,+)$. Of course, once it is a ring, then $1\in \mathbb Z$ is the unity, while $-1\in \mathbb Z$ is not, and there is only one choice for the unity, but actually then the question of this choice being natural is moot since there is only one choice at all, so it's kind of silly to say it is natural.

**Addendum** (to answer the question raised in the comment below by *unknown*)

A two dimensional vector space does not have a "natural" inner product. A two dimensional vector space *with a chosen basis* does: If $V$ is a vector space (over the field $k$) with basis $\mathbf v_1,\mathbf v_2\in V$ then one can define a "natural" inner product by $<\mathbf a,\mathbf b>:=a_1b_1+a_2b_2$ where $\mathbf a=a_1\mathbf v_1+a_2\mathbf v_2$ and $\mathbf b=b_1\mathbf v_1+b_2\mathbf v_2$. But this is only natural *after* the basis has been chosen. Basically the problem is that the definition of the inner product depends on the *choice* of the basis, so this definition is only natural if there is a natural choice of a basis (assuming there is no other structure present that could give a natural inner product; for more on this see the example of the wedge product below).

For instance if $V$ is given with a basis as above, then there is a natural choice of a basis (called the *dual basis*) for $V^*$ the dual space of $V$: Let $\phi_1: V\to k$ be defined by $\phi_1(\mathbf v_1)=1$, $\phi_1(\mathbf v_2)=0$ and extended by linearity and similarly $\phi_2: V\to k$ be defined by $\phi_2(\mathbf v_1)=0$, $\phi_2(\mathbf v_2)=1$ and extended by linearity. So, *if* you already have a chosen basis on $V$ you can find a natural choice of a basis on $V^*$ and with that you can find a natural choice of an inner product, but this will not be a natural choice if you consider the vector space without the given basis.

Otherwise, I agree, it is hard to tell what people mean by "natural". As said by many people in various answers, the essence is whether you can do the construction without making a sort of a random choice when choosing a different element would be equally good. In this example, to give an inner product you need to give a basis and unless you have some extra structure, choosing any given basis is equal to choosing any other, so the choice is non-natural.

On the other hand if there is some extra structure on the vector space then there may be a natural choice for an inner product, or more generally for a bilinear form. For instance if your vector space is a space of differential forms on a manifold, then there is no natural choice of basis, but there is a natural choice of an alternating bilinear form: the wedge product. This is actually pretty good, because this means that it is possible to define this on a manifold *locally*: picking a chart is a non-natural choice and defining the wedge product of two differential forms locally seems like it depends on a lot of choices, but it ends up being independent of these in the sense that choosing a different chart you get the same wedge product it just looks different because it is in a different basis.

**Addendum 2** (I realized that this might be an interesting comment while writing this answer to another MO question).

Here is an example of the importance of naturality: Suppose $M$ is a manifold and $U\subseteq M$ and open set. Then there is a *natural* homomorphism from the ring of regular functions on $M$ to the ring of regular functions on $U$. (If you like adjust "regular" for your favourite category; continuous, smooth, holomorphic, etc.). It can happen that this homomorphism is an isomorphism and then it has nice consequences. However, it is often important that in order to get the nice consequence one needs that *the* homomorphism induced by the embedding is an isomorphism and not that there exists *some* isomorphism. In other words, one could say that *the* natural homomorphism (=the one induced by the embedding $U\subseteq M$) is an isomorphism. For an explicit example where this matters see the above mentioned answer.

canonicalname for such X? ;) $\endgroup$ – Mark Grant Mar 1 '11 at 7:01naturalisomorphism (for instance) and 1% of them actually shownaturality, the others considering that proving it is an isomorphism is all that's really needed. I think this does not help our beloved adjective to gain much reputation. $\endgroup$ – Filippo Alberto Edoardo Jul 20 '12 at 4:58