= Solution
For a <conditional spin measurement of identical fermions> addressing the $z$ component in one spatial mode, the <Born rule> gives
$$
\boxed{P(\uparrow)=\frac12.}
$$
Conditioned on this result, the mode <spin> state is $|\uparrow_a\downarrow_b\rangle$; the other outcome produces $|\downarrow_a\uparrow_b\rangle$. The <spin> product is not a spin-exchange <eigenstate>. The <measurement> has selected an addressed mode and recorded an outcome, so this conditional <spin> description distinguishes the measured subsystem.
The full <electron> <wavefunction> still obeys <fermionic exchange symmetry>. For example, the conditional spin-up outcome in spatial mode $a$ is represented in particle coordinates by
$$
\frac1{\sqrt2}\left[\phi_a(\mathbf r_1)\phi_b(\mathbf r_2)|\uparrow_1\downarrow_2\rangle-\phi_b(\mathbf r_1)\phi_a(\mathbf r_2)|\downarrow_1\uparrow_2\rangle\right].
$$
It is antisymmetric under full particle exchange. Thus the loss of a definite symmetry of the \b[conditional <spin> product] does not mean that <measurement> turns identical <electrons> into distinguishable particles. If the outcome is discarded, the mode <spin> state is instead the equal <mixture> of the two opposite-spin products. The <measurement> must specify which spatial mode is addressed for the conditional statement to be operationally meaningful.
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