= Solution
Use <Einstein notation>, with repeated suffixes summed. The <matrix multiplication> and transpose give
$$
C_{ij}=A_{ik}(AB)_{jk}=A_{ik}A_{j\ell}B_{\ell k}.
$$
Interchange the free indices, then relabel the dummy indices $k,\ell$:
$$
C_{ji}=A_{jk}A_{i\ell}B_{\ell k}
=A_{j\ell}A_{ik}B_{k\ell}
=-A_{ik}A_{j\ell}B_{\ell k}=-C_{ij}.
$$
The negative sign uses $B_{k\ell}=-B_{\ell k}$. Therefore \b[$C$ is skew-symmetric]. No orthogonality of $A$ is needed for this part: <matrix congruence> preserves skew-symmetry.
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