Transpose-odd operator chirality in co-moving multiplicative processes: exact tail-level structure and a detectability obstruction

We ask whether transposition changes the stationary heavy tail of the co-moving recursion $\mathbf x_{t+1}=Q_tBQ_t^\top\mathbf x_t+Q_t\boldsymbolη_t$ in $d=3$. The model has independent Haar rotations, Gaussian body-frame noise with covariance $Σ$, and an explicit positive metric $\mathsf g$. The tail index $α_\star$ depends only on singular values, so $α_\star(B)=α_\star(B^\top)$. The structure theorem and polar-twist mechanism are unconditional. In the $\mathsf g=I$ chart, write $B=S+\widehat{\boldsymbolω}$: $Δ_6(B)=-16\det[\boldsymbolω,S\boldsymbolω,S^2\boldsymbolω]$. Then $B\not\sim_{O(3)}B^\top$, $Δ_6(B)\ne0$, $\operatorname{rank}[\boldsymbolω,S\boldsymbolω,S^2\boldsymbolω]=3$, and controllability of $(S,\boldsymbolω)$ are equivalent; intrinsic operator chirality is a Kalman rank condition. Unlike the exponent, the tail level need not be transpose-invariant. The exact polar-twist identity rotates the anisotropic noise frame under mirroring, showing how level asymmetry can enter. A nonzero asymmetry is established only conditionally and numerically, not by a closed-form example: under Assumption R on the SPD chart, $Δ\log C=\langle A,G\rangle+o(\|A\|)$ as $A\to0$, with $G=2H\circ[Σ,\partial_Σ\log C]$; finite-threshold simulations are consistent with this first-order law. Strict-Haar averaging makes every lagged second-order cross-moment transpose-blind; a fourth radial moment retains an odd channel. No finite scalar-weighted radial-moment combination uniformly cancels the even quadratic form while retaining the odd covector.