= Solution
Dualize the antisymmetric tensor to a vector,
$$
d_i=\frac12\varepsilon_{ijk}D_{jk},\qquad D_{ij}=\varepsilon_{ijk}d_k.
$$
An isotropic symmetric rank-two tensor depending quadratically on one vector can only be a linear combination of $\delta_{ij}d_kd_k$ and $d_id_j$: after rotating $d$ to the third axis, invariance under all rotations about that axis forces the matrix to be $\operatorname{diag}(a,a,b)$.
Since $D_{ik}D_{jk}=\delta_{ij}d^2-d_id_j$ and $D_{kl}D_{kl}=2d^2$, the most general answer may be written
$$
\boxed{T_{ijklmn}D_{kl}D_{mn}
=a\,\delta_{ij}D_{kl}D_{kl}+b\,D_{ik}D_{jk}.}
$$
The two displayed tensors are independent, for example by taking $d$ along a coordinate axis and comparing the components parallel and perpendicular to it. Hence there are exactly two independent terms.
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