= Solution
\b[The literal printed difference is zero], because the same spin-index product appears twice, and zero is invariant under every transformation. The intended <SU(2)-invariant antisymmetric spinor contraction> uses opposite indices in the second term. With $\varepsilon=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$, define $I=\phi^T\varepsilon\chi$. For any two-by-two matrix,
$$
U^T\varepsilon U=(\det U)\varepsilon.
$$
Thus for $U\in SU(2)$,
$$
\boxed{I'=(U\phi)^T\varepsilon(U\chi)=I}.
$$
Expanding the <determinant> identity proves the result without complex conjugating either <spinor>. This is the invariant alternating tensor behind the <spin singlet state>. If the two <spinors> are identical, the contraction vanishes. An antisymmetric internal two-boson state also needs an antisymmetric additional degree of freedom for overall bosonic symmetry; it does not lie in the fully symmetric fixed-two-mode sector discussed above.
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