Step 1: Determine the number of lone pairs on Xe in each compound using VSEPR theory.
Step 2: XeO$_3$
Xenon has 8 valence electrons.
It forms 3 Xe–O bonds and has 1 lone pair.
So, XeO$_3$ has 1 lone pair on Xe.
Step 3: XeF$_6$
Xenon forms 6 Xe–F bonds and has 1 lone pair.
So, XeF$_6$ also has 1 lone pair on Xe.
Step 4: Check the remaining compounds:
XeF$_2$ has 3 lone pairs.
XeF$_4$ has 2 lone pairs.
Step 5: Therefore, the pair having the same number of lone pairs is
XeO$_3$ and XeF$_6$.