= Solution
For $x'=\Lambda x$, choose the corresponding spin lift
$$
S(\Lambda)=\exp\!\left(\frac12\Omega_{\rho\sigma}M^{\rho\sigma}\right).
$$
The <spinor field> transforms as
$$
\boxed{\psi'(x')=S(\Lambda)\psi(x),\qquad
\psi'(x)=S(\Lambda)\psi(\Lambda^{-1}x).}
$$
The infinitesimal <Clifford algebra> identity from part (b) exponentiates to $S^{-1}\gamma^\mu S=\Lambda^\mu{}_{\nu}\gamma^\nu$, establishing <Lorentz covariance of the Dirac operator>. At fixed coordinates the infinitesimal variation also includes the orbital term $-\Omega^\mu{}_{\nu}x^\nu\partial_\mu\psi$.
A spatial rotation generator $M^{ij}$ is anti-Hermitian, but a <Lorentz boost> generator is Hermitian. For example,
$$
M^{01}=\frac12\gamma^0\gamma^1
=\frac12\begin{pmatrix}-\sigma^1&0\\0&\sigma^1\end{pmatrix}.
$$
Its eigenvalues are $\pm\tfrac12$. A nonzero rapidity $\zeta$ gives boost eigenvalues $e^{\pm\zeta/2}$, which are not all on the unit circle. This proves that the finite-dimensional component <Spinor representation of the Lorentz group> cannot be unitary for a positive-definite component inner product, even after a change of basis. It does not contradict a unitary action on the physical space of quantum states.
Consequently $\psi'^\dagger(x')\psi'(x')=\psi^\dagger S^\dagger S\psi$ need not equal $\psi^\dagger\psi$: it is the time component of the <Dirac current>, not a <Lorentz scalar>. The appropriate invariant form is indefinite. From $(\gamma^\mu)^\dagger=\gamma^0\gamma^\mu\gamma^0$ one obtains
$$
(M^{\rho\sigma})^\dagger\gamma^0+\gamma^0M^{\rho\sigma}=0,
\qquad S^\dagger\gamma^0S=\gamma^0.
$$
This <Dirac spinor pseudo-unitarity> gives $\bar\psi'(x')=\bar\psi(x)S^{-1}$ and hence the <Dirac scalar bilinear>
$$
\boxed{\bar\psi'(x')\psi'(x')=\bar\psi(x)\psi(x).}
$$
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