= Solution
{c}
Expand the <matrix trace> explicitly:
$$
\mathcal L=\sum_{i,j=1}^N\left[\partial_\mu\Phi_{ij}^*\partial^\mu\Phi_{ij}-m^2\Phi_{ij}^*\Phi_{ij}\right].
$$
Write $\Phi_{ij}=(x_{ij}+iy_{ij})/\sqrt2$ and collect the $2N^2$ real fields into a vector $X$. Then
$$
\mathcal L=\frac12\sum_{A=1}^{2N^2}\left[\partial_\mu X_A\partial^\mu X_A-m^2X_A^2\right].
$$
Both quadratic forms are preserved by every constant <orthogonal matrix> $R$, acting as $X\mapsto RX$. Thus the full linear internal symmetry contains $O(2N^2)$, with connected <special orthogonal group> $SO(2N^2)$ as requested. The enlargement beyond unitary transformations is possible because the real and imaginary components have identical kinetic and <mass> terms and no interactions distinguishing them.
For a <Hermitian matrix> $\Phi$, there are $N$ real diagonal entries and $N(N-1)/2$ independent complex off-diagonal entries, hence $N^2$ real fields. Choose an <orthonormal basis> $T_A$ of <Hermitian matrices> for the <trace inner product>, $\operatorname{tr}(T_AT_B)=\delta_{AB}$, and put $\Phi=\sum_AX_AT_A/\sqrt2$. The same calculation gives
$$
\boxed{SO(2N^2)\ \text{for a general complex matrix},\qquad SO(N^2)\ \text{for a Hermitian matrix},}
$$
where these are the connected rotation groups; their full orthogonal extensions are $O(2N^2)$ and $O(N^2)$. A purely quadratic theory with $m^2<0$ would have an unstable <scalar potential>, but this algebraic symmetry statement still holds.
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