I believe that an answer to the following is known but was unable to find it anywhere.
What is $\operatorname{Hilb}^2(\mathbb A^n)$?
Of course, $\operatorname{Hilb}^2(\mathbb A^1)\cong\mathbb A^2$ and $\operatorname{Hilb}^2(\mathbb A^2)\cong\mathbb A^4$, so I was thinking that probably $\operatorname{Hilb}^2(\mathbb A^n)=\mathbb A^{2n}$ could be true but I was unable to find a reference. It would be nice if someone could provide me with the answer + reference.
For each $n$ there is a natural map $$ Hilb^2(\mathbb{A}^n) \to Gr(2,n+1) \setminus Gr(2,n) $$ that takes a subscheme to the unique line supporting it (considered as a line in $\mathbb{P}^n$ that is not contained in $\mathbb{P}^{n-1} = \mathbb{P}^n \setminus \mathbb{A}^n$). All fibers of this map can be identified with $Hilb^2(\mathbb{A}^1) = \mathbb{A}^2$. Moreover this map is a locally trivial vector bundle (the restriction of the symmetric square of the tautological bundle from the Grassmannian). In particular, it is not isomorphic to $\mathbb{A}^{2n}$ for $n \ge 2$ (even for $n = 2$!).