 Solution

Independent global phase symmetries would rotate $\varphi \mapsto e^{-i\alpha }\varphi$ and $\Phi \mapsto e^{-i\beta }\Phi$. The cubic interaction vertex preserves only $2\alpha +\beta =0$ modulo $2\pi$. The continuous internal symmetry of a classical field theory is therefore the circle groupNormalize the Noether charge so that $q_{\varphi }=1$. Then $q_{\Phi }=-2$, $q_{\bar{\varphi }}=-1$ and $q_{\bar{\Phi }}=2$. The Noether current iswhere $A\overleftrightarrow{\partial }B=A\partial B-(\partial A)B$. For example, the Euler-Lagrange field equations givewhich cancel with the relative coefficient $-2$. The quartic interaction vertex separately preserves both original phase rotations.

There is also a discrete charge conjugation symmetry $(\varphi ,\Phi )\mapsto ({\varphi }^{†},{\Phi }^{†})$ for these real couplings. Including it, the internal symmetry is $U(1)⋊{\mathbb{Z}}_2\cong O(2)$, with charge conjugation reversing the Noether charge. The connected symmetry and its conserved current are the $U(1)$ above.