Assume that two smooth quasi-projective vareities $X,Y$ have the same class in the Grothendieck ring of varieties.



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> Do the symmetry products $X^{(n)}$ and $Y^{(n)}$ also have the same class?
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Moreover, if there are some feasible criteria to judge if two varieties have the same class?


Edit: I leave this part of original the question as well as the answer as a complete closed post.



Symmetry product in Grothendieck ring of varieties
This is just a quick explanation. Let $k$ be an algebraically closed field. Let $\mathcal{Var}\_k$ denote the set of (embedded) quasi-projective $k$-varieties. The Grothendieck group of varieties, $K\_0(\mathcal{Var}\_k)$, is the quotient of the free Abelian group on $\mathcal{Var}\_k$ by all relations, $$[X] = [U]+[C],$$ if $C$ is $k$-isomorphic to a closed subvariety of $X$ whose open complement is $k$-isomorphic to $U$. This is actually a commutative ring under the product $[X]\cdot [Y] = [X\times\_{\text{Spec}\ k} Y]$. The identity element is $1=[\text{Spec}\ k]$.


Inside the power series ring $K\_0(\mathcal{Var}\_k)[[t]]$, the subset of power series whose constant coefficient equals $1$ is an Abelian group under multiplication; call this group $G$.  

For every $X$ in $\mathcal{Var}\_k$, Kapranov defines an element in $G$, the **motivic zeta function**, $$\zeta\_X(t) = 1 + \sum\_{n\geq 1}[\text{Sym}^n(X)]t^n.$$
For every integer $n\geq 1$, there is a partition of $\text{Sym}^n(X)$ into locally closed subsets $Z\_{n,m}$ for $m=0,\dots,n$, where $Z\_{n,m}$ denotes the subvariety parameterizing length $n$ zero-cycles on $X$ whose intersection with $U$ has length precisely $m$. Then the "residual" zero-cycle is a length $n-m$ zero-cycle supported on $C$. In this way we get a $k$-isomorphism, $$Z\_{n,m} \cong \text{Sym}^m(U)\times\_{\text{Spec}\ k}\text{Sym}^{n-m}(C).$$
This is precisely the same as the identity, $$\zeta\_X(t) = \zeta\_U(t)\cdot \zeta\_C(t).$$ Thus, the zeta function from $\mathcal{Var}\_k$ to the Abelian group $G$ satisfies the relations to factor through a group homomorphism from $K\_0(\mathcal{Var}\_k)$, $$\zeta:K\_0(\mathcal{Var}\_k) \to (1+tK\_0(\mathcal{Var}\_k)[[t]])^\times.$$ In particular, the motivic zeta function $\zeta\_X(t)$
only depends on the class of $X$ in $K\_0(\mathcal{Var}\_k)$. Consequently, for every integer $n\geq 1$, the class $[\text{Sym}^n(X)]$ in $K\_0(\mathcal{Var}\_k)$ only depends on the class of $X$ in $K\_0(\mathcal{Var}\_k)$.


