 Solution

Choose a normalization in which the circle group has period $2\pi$ and the U(1) charges ${\hat{Q}}_i$ of the complex scalar fields are integers. By restriction of a representation,Write $v_i=⟨{\varphi }_i⟩$. A vacuum expectation value $v_i\ne 0$ is preserved by $e^{i\theta }$ precisely when $e^{i{\hat{Q}}_i\theta }=1$. Therefore the specified finite cyclic group survives spontaneous symmetry breaking exactly when every condensing complex scalar field obeysOnly U(1) charges of fields with nonzero vacuum expectation values enter this test; in the paper all $N$ fields condense.

Suppose at least one condensing U(1) charge is nonzero and put $d=\gcd (|{\hat{Q}}_i|:v_i\ne 0)$. Since each such U(1) charge is a multiple of $d$, every $d$th root of unity preserves the vacuum expectation values. Conversely, the Bezout identity expresses $d$ as an integer combination of these U(1) charges, so a common stabilizer obeys $e^{id\theta }=1$. The discrete remnant of a circle symmetry is consequentlyA nontrivial retained finite cyclic group exists exactly when $d>1$, and its possible orders are the divisors $n>1$ of $d$. For example, condensing U(1) charges $6$ and $10$ leave ${\mathbb{Z}}_2$. If $d=1$, the stabilizer is trivial. If all condensing U(1) charges vanish, the entire circle group is unbroken and there is no largest finite $n$; this case does not break the symmetry assumed in the question.