What Happens If an Unbroken Flavor Symmetry Exists?
  Without assuming any specific flavor symmetry and/or any specific mass matrix forms, it is demonstrated that if a flavor symmetry (a discrete symmetry, a U(1) symmetry, and so on) exists, we cannot obtain the CKM quark mixing matrix $V$ and the MNS lepton mixing matrix $U$ except for those between two families for the case with the completely undegenerated fermion masses, so that we can never give the observed CKM and MNS mixings. Only in the limit of $m_{\nu 1} =m_{\nu 2}$ ($m_d=m_s$), we can obtain three family mixing with an interesting constraint $U_{e3}=0$ ($V_{ub}=0$). 
Yoshio Koide