= Solution
Independent <global phase symmetries> would rotate $\phi\mapsto e^{-i\alpha}\phi$ 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 group>
$$
\boxed{U(1):\quad\phi\mapsto e^{-i\alpha}\phi,\qquad\Phi\mapsto e^{2i\alpha}\Phi.}
$$
Normalize the <Noether charge> so that $q_\phi=1$. Then $q_\Phi=-2$, $q_{\bar\phi}=-1$ and $q_{\bar\Phi}=2$. The <Noether current> is
$$
\boxed{j^\mu=i\phi^\dagger\overleftrightarrow{\partial^\mu}\phi-2i\Phi^\dagger\overleftrightarrow{\partial^\mu}\Phi,\qquad\partial_\mu j^\mu=0,}
$$
where $A\overleftrightarrow\partial B=A\partial B-(\partial A)B$. For example, the <Euler-Lagrange field equations> give
$$
\partial_\mu(i\phi^\dagger\overleftrightarrow{\partial^\mu}\phi)=ih(\phi^2\Phi-\phi^{\dagger2}\Phi^\dagger),\qquad
\partial_\mu(i\Phi^\dagger\overleftrightarrow{\partial^\mu}\Phi)=\frac{ih}{2}(\phi^2\Phi-\phi^{\dagger2}\Phi^\dagger),
$$
which cancel with the relative coefficient $-2$. The quartic <interaction vertex> separately preserves both original phase rotations.
There is also a discrete <charge conjugation> symmetry $(\phi,\Phi)\mapsto(\phi^\dagger,\Phi^\dagger)$ for these real couplings. Including it, the internal symmetry is $U(1)\rtimes\mathbb Z_2\cong O(2)$, with <charge conjugation> reversing the <Noether charge>. The connected symmetry and its <conserved current> are the $U(1)$ above.
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