= Solution
\b[True.] Since $A$ and $A'$ are antisymmetric, the matrix $D=R^TA'R-A$ is antisymmetric. Invariance of contractions with every <antisymmetric second-rank tensor> gives $D:B=0$ for every antisymmetric test matrix $B$. Such a matrix can be freely prescribed in one basis and then transformed to define a valid test <tensor>. Choose $B=D$ in that basis. Then
$$
D:D=\sum_{ij}D_{ij}^2=0,
$$
so $D=0$ and $A'=RAR^T$. Together with the given antisymmetry, this proves that $A$ is an <antisymmetric second-rank tensor>. The <antisymmetric contraction test for an antisymmetric tensor> works because the <Frobenius inner product> is nondegenerate on the antisymmetric subspace.
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