Let $M_{ij}\in\mathbb R^{n\times n}$ be the matrix with $1$ on position $(i,j)$ and $0$ elsewhere and let $V\subset\mathbb R^{n\times n}$ be the subspace of antisymmetric matrices. Then the "standard" isomorphism between $\mathbb R^n\wedge\mathbb R^n$ and $V$ maps $e_i\wedge e_j$ to $M_{ij}-M_{ji}$ (this was explained here).
I guess that this isomorphism is in some sense unique, i.e. this a "natural" isomorphism, but I don't know what "natural" means in this context. Can someone explain?
Natural here has a precise meaning in the language of categories but that may not be appropriate here. Instead you can think of it as meaning "coordinate independent." This seems a little weird because the map you defined is written in terms of coordinates but you actually don't need to do this.
The coordinate-free definition of the map looks like this. Define a map $\varphi:\mathbb R^n\wedge\mathbb R^n\to V$ by $$\varphi(v\wedge w)(u)=(v\cdot u)w-(w\cdot u)v.$$ If you write this map in coordinates you'll see that it's actually the same as the one you defined. The advantage to my definition is that I make no use of the fact that we're in $\mathbb R^n.$ This definition works the same way for every inner product space $V,$ and this is what earns it the descriptor "natural."