= Solution
{c}
Let $X=\Phi^\dagger\Phi$. Under the global transformation of the <chiral matrix scalar model>,
$$
\Phi'=e^{i\theta}U_L\Phi U_R^\dagger,\qquad X'=U_RXU_R^\dagger.
$$
The phase and $U_L^\dagger U_L$ cancel, and $U_R^\dagger U_R=\mathbf1$. The <cyclic property of the trace> implies
$$
\operatorname{tr}X'=\operatorname{tr}X,\qquad
\operatorname{tr}(X'^2)=\operatorname{tr}(U_RX^2U_R^\dagger)=\operatorname{tr}X^2.
$$
Therefore each of the three terms in the <scalar potential> is invariant separately. Since $\theta,U_L,U_R$ are constant, $\partial_\mu\Phi'=e^{i\theta}U_L(\partial_\mu\Phi)U_R^\dagger$, giving invariance of the kinetic <matrix trace> by the same argument. Hence
$$
\boxed{G=SU(N)_L\times SU(N)_R\times U(1)}
$$
is a global symmetry group of the <Lagrangian density>. Its action has a finite central kernel, which does not affect its <Lie algebra> or the broken-generator counting below. For special parameter choices the symmetry can be larger; the displayed group is a symmetry for the stated potential without such extra assumptions.
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