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True. Since $A$ and $A^{\prime }$ are antisymmetric, the matrix $D=R^T{A}^{\prime }R-A$ 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. Thenso $D=0$ and $A^{\prime }=RA{R}^T$. 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.