= Solution
Divide the infinitesimal <variation> by its constant parameter $\alpha$, so the transformation generator is $(\delta\phi_1,\delta\phi_2)=(\phi_2,-\phi_1)$. The variation of the <kinetic term> is
$$
\partial_\mu\phi_1\partial^\mu\phi_2-\partial_\mu\phi_2\partial^\mu\phi_1=0,
$$
and that of $\phi_1^2+\phi_2^2$ is $2\phi_1\phi_2-2\phi_2\phi_1=0$. Thus the <Lagrangian density> is invariant pointwise, without a nonzero boundary current. The finite transformation is a rotation of the two <real scalar fields>, giving an <SO(2)> <global symmetry in field theory>.
<Noether's theorem> supplies the parameter-independent <Noether current>
$$
\boxed{j^\mu=\phi_2\partial^\mu\phi_1-\phi_1\partial^\mu\phi_2.}
$$
Its <divergence> is $\partial_\mu j^\mu=\phi_2\Box\phi_1-\phi_1\Box\phi_2$, since the first-derivative terms cancel. Both fields obey the same <Klein-Gordon equation>, so this vanishes on solutions. With spatial boundary conditions giving zero flux at infinity, the <Noether charge> is
$$
\boxed{Q=\int d^3x\,(\phi_2\dot\phi_1-\phi_1\dot\phi_2),\qquad \dot Q=0.}
$$
A finite box with periodic <boundary conditions> works as well. The spatial integral is conserved because the local <continuity equation> converts its time derivative into a boundary flux; neither vanishing local current nor constant individual fields is required. Equivalently, $\Phi=(\phi_1+i\phi_2)/\sqrt2$ transforms as $\Phi\mapsto e^{-i\alpha}\Phi$, exhibiting the equivalent <global U(1) symmetry> and fixing the sign of the generator.
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