Cyclobutane and its substituted derivatives readily undergo a ring flip (or ring inversion) as pictured below.
The barrier to ring flipping is very low, around 1.5 kcal/mole. So at room temperature the flipping process is happening very rapidly. Cyclobutane exists as a puckered molecule as depicted in $\ce{A}$ and $\ce{B}$. In the ring flipping process, the molecule passes through a transition state where the cyclobutane ring is planar.
Only molecules that belong to symmetry classes (point groups)
- $\ce{C_{n}}$ (the molecule only contains a $\ce{C_{n}}$ axis)
- $\ce{C_{nv}}$ (the molecule contains a $\ce{C_{n}}$ axis and a $\ce{\sigma}_{v}$ plane)
- $\ce{C_{s}}$ (the molecule only has a plane of symmetry)
can have a dipole moment.
Conformers $\ce{A}$ and $\ce{B}$ both have $\ce{C_{s}}$ symmetry (the symmetry plane passes through the two cyclobutane carbons bearing the substituents) and therefore do have a dipole moment; however, the planar transition state does not belong to one of these point groups (it has $\ce{C_{2h}}$ symmetry) and therefore does not have a dipole moment.
On average, when flipping is rapid, the dipole moments of conformers $\ce{A}$ and $\ce{B}$ average out to zero; or said differently, the molecule "looks like" the planar transition state which does not have a dipole moment. So at room temperature where flipping is rapid, the molecule does not display a dipole moment. If you cooled the system down to a very low temperature where conformers $\ce{A}$ and $\ce{B}$ were not flipping, then you could measure a dipole moment.