= Solution
\b[The nine normal coordinates transform as $A_{1g}+B_{1g}+B_{2g}+A_{2u}+B_{2u}+2E_u$.] Place the molecule in the $xy$ plane, with its four outer atoms on the $x$ and $y$ axes, as specified for the vertical reflection planes. In the class order
$$
E,\ 2C_4,\ C_2,\ 2C'_2,\ 2C''_2,\ i,\ 2S_4,\ \sigma_h,\ 2\sigma_v,\ 2\sigma_d,
$$
the fifteen Cartesian displacements have <characters of a representation>
$$
\chi_{3N}=(15,1,-1,-3,-1,-3,-1,5,3,1).
$$
For each operation this character is the number of fixed atoms times the trace of its polar-vector matrix. Subtract <translations> $A_{2u}+E_u$ and <rotations> $A_{2g}+E_g$, giving
$$
\chi_{\rm vib}=(9,-1,1,-1,1,-3,-1,5,3,1).
$$
<Molecular character reduction> in the <D4h molecular point group> yields the stated decomposition. Its dimensions sum to $1+1+1+1+1+2(2)=9=3(5)-6$, providing an independent degree-of-freedom check.
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