Given a smooth algebraic curve $C$ with a closed point $p$. The symmetric product $\text{Sym}^i(C)$ has a closed subvariety $\text{Sym}^{i-1}(C)$ which its embedding is given by adding the point $p$. Let $\text{Spec}(A)$ be an affine in $C$ containing $p$. Is the $n$-th nilpotent thickening of $\text{Sym}^{i-1}(C)$ in $\text{Sym}^i(C)$ a scheme over $\text{Spec}(A/m^{n+1})$ where $m$ is the maximal/prime ideal corresponding to $p$?
Note that zeroth nilpotent thickening is the subscheme itself, the first nilpotent is given by second powers and so on.
This problem seems trivial intuitively, since symmetric product is the quotient of product, if we look at the product and the inverse image of $\text{Sym}^{i-1}(C)$, its $n$-th nilpotent thickening is the union of bunch of $\text{Spec}(A/m^{n+1})$-schemes (each irreducible component). It seems that gluing these schemes (The ideal corresponding to the gluing is going to be the product of ideals) will have a natural $\text{Spec}(A/m^{n+1})$-scheme structure. There also seems to be symmetry to this $\text{Spec}(A/m^{n+1})$-scheme structure that descends to the symmetric product.
Question: "Is the n-th nilpotent thickening of $Sym^{i−1}(C)$ in $Sym^i(C)$ a scheme over $Spec(A/m^{n+1})$ where $m$ is the maximal/prime ideal corresponding to p?"
Answer: Note that the symmetric product is not functorial for prime ideals as you claim if you use the symmetric group to define the symmetric product:
If $k$ is the base field and $C$ is a curve over $k$, let $U:=Spec(A) \subseteq C$ be an affine open subscheme. Let us define the symmetric product locally using the following construction: Define
$$A_i:=A^{\otimes_k i}:=A\otimes_k A\otimes \cdots \otimes_k A.$$
There is a canonical action of the symmetric group $S_i$ on $i$ elements on $A_i$, and by definition
$$ Sym^i(U):= Spec(A_i^{S_i})$$
is the spectrum of the invariant ring $A_i^{S_i}$. Given an arbitrary closed point $\mathfrak{m} \subseteq A$ you get a canonical map
$$ U^{\times_k (i-1)} \times_k Spec(\kappa(\mathfrak{m})) \rightarrow U^{\times_k i}$$
and if $\mathfrak{m}$ is a $k$-rational point you get a canonical map
$$i_{\mathfrak{m}}: U^{\times_k (i-1)} \rightarrow U^{\times_k i}$$
by "adding the point $\mathfrak{m}$", but this map does not induce a map of symmetric products for general points as you claim:
Example: Let $A:=k[x]$ with $A_3:=A^{\otimes_k 3}\cong k[x_1,x_2,x_3]$ and let $\mathfrak{m}:=(x_3-a_3)\subseteq k[x_3]$ with $a_3\in k$ be a $k$-rational point. It follows
$$A_3^{S_3}\cong k[s_1,s_2,s_3]$$
where $s_1:=x_1+x_2+x_3, s_2:=x_1x_2+x_1x_3+x_2x_3, s_3:=x_1x_2x_3$
and $A_2^{S_2}\cong k[t_1,t_2]$ with
$$t_1:=x_1+x_2, t_2:=x_1x_2$$
and it follows the map
$\phi: A_3 \rightarrow A_2$ defined by $\phi(x_3):=a_3$ maps as follows:
$$\phi(s_1):=x_1+x_2+a_3:= t_1+a_3$$
$$\phi(s_2):= a_3t_1+t_2$$
$$\phi(s_3):= a_3t_2,$$
hence you do get an induced map of symmetric products
$$\phi^*: Sym^2(U) \rightarrow Sym^3(U)$$
in the case when your point $\mathfrak{m}$ is $k$-rational. You do not get an induced map $\phi^*$ for a general point:
Example: Let $k$ be the real numbers and let $K$ be the complex numbers. Let $(p(x)) \subseteq k[x_3]$ be a maximal ideal with residue field $K$ (ie we map $x_3$ to $z_3\in K$). In the above construction you get a map
$$k[x_1,x_2,x_3] \rightarrow k[x_1,x_2]\otimes_k K \cong K[x_1,x_2]$$
and an induced map
$$k[x_1,x_2,x_3]\rightarrow K[x_1,x_2]$$
inducing a map
$$k[s_1,s_2,s_3] \rightarrow K[t_1,t_2]$$
and
$$Sym^2(K\times_k U) \rightarrow Sym^3(U).$$
Hence for an arbitrary point $p\in C$, you may not view $Sym^{i-1}(C)$ as a closed subscheme of $Sym^i(C)$ by "adding the point $p$": There is a canonical map
$$ Sym^2(K\times_k U) \rightarrow Sym^2(U)$$
but no closed immersion
$$Sym^2(U) \rightarrow Sym^3(U)$$
in general.