0\}\cup\{0 \}$ where as previously, if $J=(j_n)$, then $j_{n(J)}$ is the last non-zero coordinate of $J$. Thus, $\sgn_\od(\chi_J)=\sgn(j_{n(J)})$. Observe that if $\epsilon=(\epsilon_n)\in\{-1,1\}^\N$, then the set $\od(\epsilon)= \{J=(j_n)\in\dualz:\epsilon_{n(J)}j_{n(J)}>0\}\cup\{0 \}$ is also an order on $\dualz$. In this case, $\sgn_{\od(\epsilon)}(\chi_J)=\epsilon_{n(J)}\sgn(j_{n(J)})$. We now state a simple identity that links the unconditionality of martingale difference sequences to harmonic conjugation with respect to orders: for every $n\geq 1$, and all $J=(j_n)\in\Z^n\setminus\Z^{n- 1}$,\ we have $$\epsilon_n =\sgn_{\od(\epsilon)}(\chi_J)\, \sgn_{\od}(\chi_J).\qquad \tag{3.2}$$ To verify (3.2), simply note that for each $n\in\N$, if $J\in\Z^n\setminus\Z^{n-1}$, then $n(J)=n$. From (3.2), one immediately obtains that $$ T_P \circ T_{P(\epsilon)} \bigg( \sum_{k=1}^n \sum_{J\in K_k} a_J\,\chi_J \bigg) = \sum_{k=1}^n\epsilon_k\bigg(\sum_{J\in K_k}a_J\,\chi_J \bigg), \tag{3.3} $$ which expresses the martingale transform on the right side as a composition of two conjugate function operators. Applying Theorem~2.1 twice yields (3.1) and implies our next and last result. \proclaim{(3.1) Theorem} Suppose $X$ is a Banach space, and let $1