This problem is highly related to this one and in fact it is the same question applied to a very specific situation.
Given a smooth projective curve $C$, let $\text{Sym}^i(C)$ be the $i$-th symmetric product. There is a closed immersion $\text{Sym}^{i-1}(C)\hookrightarrow \text{Sym}^i(C)$ by adding a fixed point. Let $Z^{(n)}$ be the $n$-th nilpotent thickening of this immersion. My question is: does the colimit of $Z^{(n)}$'s exist in the category of schemes and is it smooth?
The first step would be to lift the problem to the product of curves. Is the colimit in the category of schemes of infinitesimal neighborhoods of $C^{i-1}$ in $C^{i}$ equal to $C^{i-1}\times \text{Spec}(k[[t]])$? This is true and follows from Grothendieck's algebraization. This post explains the colimit problem in the category of schemes and it is easy to see that $C^{i-1}\times \text{Spec}(k[[t]])$ is the colimit of nilpotent thickenings of $C^{i-1}$ in $C^i$.
Now the problem is that how to glue these together. Is it possible to glue $i$-copies of $C\times \ldots \times C\times \text{Spec}(k[[t]]) \times \ldots \times C $ where for each $i$ the $i$-th component is $\text{Spec}(k[[t]])$ in the category of schemes? (At first I thought this is obvious but this is neither gluing along open nor closed subschemes). If it is possible to glue these then we need to take quotient by the symmetric group.
If the answer to the problem is positive, is the colimit smooth?
