= Solution
Replacing X-rays by neutrons of the same <wavelength> keeps the geometric Bragg positions, but replaces electron-density amplitudes by nuclear coherent scattering lengths. For Al all sites still carry the same nonzero amplitude, so its six reflections remain. For Si the identical two-site basis still has the same destructive-interference conditions, so its four reflections remain. Isotopic randomness can add diffuse/incoherent scattering but does not create a new periodic basis with different systematic Bragg absences.
K and Cl, however, have different nuclear amplitudes despite their isoelectronic ionic <electron> clouds. For natural isotopic composition, $b_K=3.67$ fm and $b_{\rm Cl}=9.577$ fm. The all-odd/all-even basis-intensity ratio becomes
$$
\frac{|b_K-b_{\rm Cl}|^2}{|b_K+b_{\rm Cl}|^2}=0.199,
$$
which is no longer the strong X-ray cancellation. The two odd rings should therefore be observable with suitable sensitivity, giving
$$
\boxed{N_{\rm Al}=6,\qquad N_{\rm KCl}=6,\qquad N_{\rm Si}=4}.
$$
The numerical contrast uses the NIST bound coherent lengths.<Neutron scattering lengths and cross sections by NIST>[Complete table, natural-element K and Cl rows: coherent b in femtometres; also Al and Si rows]{ref} Counts here mean allowed and reasonably observable rings in the specified range, assuming sufficient angular coverage and resolution; no theory alone fixes an instrument's intensity threshold. These nonmagnetic examples use nuclear elastic scattering, not magnetic scattering.
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