= Solution
The continuous symmetry is
$$
G=U(N)_L\times U(N)_R,
\qquad
M\longmapsto LMR^\dagger.
$$
Strictly, the diagonal central $U(1)$ acts trivially, so the faithful group is the quotient by that common center. Since $L$ and $R$ are <unitary matrices>,
$$
\partial_\mu M\longmapsto L(\partial_\mu M)R^\dagger,
\qquad
M^\dagger M\longmapsto R(M^\dagger M)R^\dagger.
$$
Every term is therefore unchanged by cyclicity of the <matrix trace>:
$$
\operatorname{Tr}(\partial^\mu M^\dagger\partial_\mu M)
\longmapsto
\operatorname{Tr}[R(\partial^\mu M^\dagger\partial_\mu M)R^\dagger],
$$
and the same conjugation argument applies to $\operatorname{Tr}(M^\dagger M)$ and $\operatorname{Tr}[(M^\dagger M)^2]$.
Solved by gpt-5.6-sol high.
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