In order for $\overline{\psi} \psi$ to be a Lorentz scalar, we should have $$\lambda^{-1} = \lambda^{\dagger}\gamma^0$$ or equivalently $$\lambda^{\dagger}\gamma^0\lambda = 1$$ what else do you need to know?

The non-unitary is caused by $\gamma^0$ in $\lambda^{\dagger}\gamma^0\lambda = 1$. Without $\gamma^0$, $\lambda$ as in $$\lambda^{\dagger}\lambda = 1$$ would be unitary.

In order for $\overline{\psi} \psi$ to be a Lorentz scalar, we should have $$\lambda^{\dagger}\gamma^0\lambda = \gamma^0$$ or equivalently (assuming $(\gamma^0)^2 = 1$) $$\lambda^{-1} = \gamma^0\lambda^{\dagger}\gamma^0$$ what else do you need to know?

The non-unitarity is caused by $\gamma^0$ in $\lambda^{\dagger}\gamma^0\lambda = \gamma^0$. Without $\gamma^0$, $\lambda$ as in $$\lambda^{\dagger}\lambda = 1$$ would be unitary.