Consider two multivariate polynomials with integer coefficients $P(x_1,\ldots,x_n)$ and $Q(x_1,\ldots,x_n)$.
Let's evaluate the two polynomials over all the permutations of $(x_1,\ldots,x_n)$ and multiply them together to get the simmetric polynomials $P^*(x_1,\ldots,x_n)$ and $Q^*(x_1,\ldots,x_n)$ respectively. For example with three variables we have $P^*(x_1,x_2,x_3)=P(x_1,x_2,x_3)P(x_1,x_3,x_2)P(x_2,x_1,x_3)P(x_2,x_3,x_1)P(x_3,x_1,x_2)P(x_3,x_2,x_1)$.
Is it true that:
$$P^*(x_1,\ldots,x_n) = Q^*(x_1,\ldots,x_n)$$
implies that $P(x_1,\ldots,x_n)$ is equal to $Q(x_1,\ldots,x_n)$ or $Q$ evaluated on a permutation of $(x_1,\ldots,x_n)$?
Even with this fix it's still not true. If $\sigma \in S_n$ is a permutation and $P$ is a polynomial write $\sigma(P)$ for the result of the permutation $\sigma$ applied to the polynomial $P$. The reason is that we could have two polynomials $F, G$ such that $P = FG$ while $Q = \sigma_1(F) \sigma_2(G)$ for two different permutations $\sigma_1, \sigma_2$. For an explicit example, take $n = 2, P = x_1^2, Q = x_1 x_2$.
However, the statement is true if $P$ and $Q$ are irreducible; in this case you can just consider the irreducible factorization of $P^{\ast}$ and of $Q^{\ast}$ to conclude that $P$ is a permutation of $Q$ up to possibly a multiplication by $-1$ if $n$ is even (e.g. we could have $n = 2, P = x_1, Q = -x_2$).