There's just the one (up to symmetric monoidal equivalence).

For this answer, I'll need a lemma: if $C$, $D$ are $Vect$-enriched categories and $F, G$ are additive $Vect$-enriched functors $C \to D$, then any natural transformation $F \to G$ is $Vect$-enriched natural. This is a consequence of the fact that the underlying-set functor $\hom(k, -): Vect \to Set$ (where $k$ is the ground field) is faithful; see Kelly's Basic Concepts of Enriched Category Theory, bottom of page 10.

Let $\otimes_{bil}: Vect \times Vect \to Vect$ be the tensor product. (All tensor products are over the ground field.) The subscript $bil$ indicates that as it stands, the tensor product is not $Vect$-enriched, but at least it is "locally bilinear" in the sense that the structure maps

$$\hom(V, V') \times \hom(W, W') \to hom(V \otimes W, V' \otimes W')$$

are bilinear. Similarly, let $\sigma_{bil}: Vect \times Vect \to Vect \times Vect$ denote the obvious transposition functor, taking a pair $(V, W)$ to $(W, V)$. A symmetry structure on $Vect$ is, by definition, an invertible natural transformation of the form

$$\otimes_{bil} \to \otimes_{bil} \circ \sigma_{bil}$$

satisfying some extra coherence properties.

There is an obvious passage from bilinearity to linearity, where we replace $Vect \times Vect$ by $Vect \otimes Vect$ (objects of $Vect \otimes Vect$ are again pairs $(V, W)$, but $\hom((V, V'), (W, W'))$ is the tensor product $\hom(V, V') \otimes \hom(W, W')$ instead of the cartesian product). We have corresponding functors $\otimes: Vect \otimes Vect \to Vect$, $\sigma: Vect \otimes Vect \to Vect \otimes Vect$, and the invertible natural transformations above are in natural bijection with invertible natural transformations of the form

$$\otimes \to \otimes \circ \sigma$$

which are, by the lemma, $Vect$-enriched.

In short, we are trying to understand $Vect$-enriched natural transformations of the form

$$V \otimes W \to W \otimes V$$

or equivalently, extranatural transformations of the form

$$V \to \hom(W, W \otimes V) \cong \hom(\hom(k, W), W \otimes V).$$

By the enriched Yoneda lemma, these are in natural bijection with enriched natural transformations of the form

$$V \to k \otimes V \cong V.$$

Applying the Yoneda trick a second time, these are in natural correspondence with morphisms $k \to k$, i.e., scalars. Since the transformation $V \otimes W \to W \otimes V$ is to be invertible, the corresponding scalar $\lambda \in k$ is also invertible.

Additionally, the required coherence condition puts another constraint on the scalar: if you work through the hexagon identity, you find out that $\lambda^2 = \lambda$. Therefore $\lambda = 1$, meaning that the only symmetry structure is the standard one.

A key to the argument is the faithfulness of $\hom(k, -): Vect \to Set$. This gives a clue that to get more interesting examples where there is a variety of possible symmetric monoidal structures, you might pass to an enriched category context where the functor $hom(I, -): V \to Set$ represented by the monoidal unit $I$ is *not* faithful, as was the case with $V$ the category of $\mathbb{Z}_2$-graded vector spaces, mentioned in an answer to another question of yours.