 Solution

For the chiral transformation $\psi \mapsto e^{i\alpha {\gamma }^5}\psi$, one also has $\bar{\psi }\mapsto \bar{\psi }{e}^{i\alpha {\gamma }^5}$. The massless kinetic and vector-current terms are invariant because $\{{\gamma }^5,{\gamma }^{\mu }\}=0$. Since ${\gamma }^5$ commutes with $[{\gamma }^{\mu },{\gamma }^{\nu }]$, the Pauli bilinear transforms with $e^{2i\alpha {\gamma }^5}$ and is not invariant. Thus the continuous classical axial symmetry requires $\lambda =0$.

A fermion mass also breaks the symmetry, so in the massive theory one additionally needs $m=0$, which contradicts a genuinely massive fermion. Making $\alpha$ local produces $({\partial }_{\mu }\alpha )\bar{\psi }{\gamma }^{\mu }{\gamma }^5\psi$. No choice of the vector coupling $e$ or Pauli coupling $\lambda$ cancels this term; gauging it requires an axial gauge field, and at the quantum level one must also address the chiral anomaly.