= Solution
Use <gamma matrices> satisfying the <Clifford algebra> relation $\{\gamma^\mu,\gamma^\nu\}=2\eta^{\mu\nu}$, with $\gamma^{0\dagger}=\gamma^0$ and $\gamma^{i\dagger}=-\gamma^i$. The <Dirac adjoint> is $\bar\Psi=\Psi^\dagger\gamma^0$. Take the <Dirac action>
$$
S_D=\int d^4x\,\bar\Psi(i\gamma^\mu\partial_\mu-m)\Psi.
$$
It differs from the manifestly Hermitian form with $(i/2)\bar\Psi\gamma^\mu\overleftrightarrow\partial_\mu\Psi$ only by a boundary term. Treat $\Psi$ and $\bar\Psi$ as independent variables when applying the <principle of stationary action>. Varying $\bar\Psi$ gives
$$
\boxed{(i\gamma^\mu\partial_\mu-m)\Psi=0.}
$$
Varying $\Psi$ and integrating by parts gives the <adjoint Dirac equation>, $i(\partial_\mu\bar\Psi)\gamma^\mu+m\bar\Psi=0$. Multiplying the <Dirac equation> by $i\gamma^\nu\partial_\nu+m$ also gives $(\Box+m^2)\Psi=0$, so its dispersion relation is <Lorentz invariant>.
For a <Lorentz transformation> $x'=\Lambda x$, the field transforms in the <Spinor representation of the Lorentz group>:
$$
\Psi'(x')=S(\Lambda)\Psi(x),\qquad S^{-1}\gamma^\mu S=\Lambda^\mu{}_{\nu}\gamma^\nu.
$$
Since $\partial'_\mu=(\Lambda^{-1})^\nu{}_{\mu}\partial_\nu$, this identity gives the <Lorentz covariance of the Dirac operator>:
$$
(i\gamma^\mu\partial'_\mu-m)\Psi'(x')=S(\Lambda)(i\gamma^\nu\partial_\nu-m)\Psi(x).
$$
Thus every solution is carried to another solution. The field is a spinor rather than a <four-vector>; the transformation of the <gamma matrices> supplies the necessary covariance. The <Dirac action> is <Lorentz invariant> because its integrand is a scalar and $d^4x$ is invariant.
For electric charge $q$, use the <gauge covariant derivative> $D_\mu=\partial_\mu+iqA_\mu$. The <minimal electromagnetic coupling of a Dirac field> is
$$
\mathcal L=\bar\Psi(i\gamma^\mu D_\mu-m)\Psi-\frac14F_{\mu\nu}F^{\mu\nu}=\bar\Psi(i\gamma^\mu\partial_\mu-m)\Psi-qA_\mu\bar\Psi\gamma^\mu\Psi-\frac14F_{\mu\nu}F^{\mu\nu}.
$$
The local <gauge transformations> in this convention are
$$
\boxed{\Psi'=e^{-iq\alpha(x)}\Psi,\qquad\bar\Psi'=\bar\Psi e^{iq\alpha(x)},\qquad A'_\mu=A_\mu+\partial_\mu\alpha.}
$$
Direct substitution gives $D'_\mu\Psi'=e^{-iq\alpha}D_\mu\Psi$, which proves <gauge covariance of the charged Dirac equation>. The mass and kinetic terms are invariant, and $F'_{\mu\nu}=F_{\mu\nu}$, so the full action has <gauge invariance>. The <conserved current> is $j^\mu=q\bar\Psi\gamma^\mu\Psi$. It transforms as a <four-vector> under <Lorentz transformations>, and varying $A_\mu$ gives $\partial_\nu F^{\nu\mu}=j^\mu$.
To make the infinitesimal spinor transformation explicit, write $\Lambda^\mu{}_{\nu}=\delta^\mu{}_{\nu}+\omega^\mu{}_{\nu}$ with $\omega_{\mu\nu}=-\omega_{\nu\mu}$. Define
$$
\sigma^{\mu\nu}=\frac i2[\gamma^\mu,\gamma^\nu],\qquad S=1-\frac i4\omega_{\mu\nu}\sigma^{\mu\nu}+O(\omega^2).
$$
The <Lorentz generators from gamma-matrix commutators> have precisely the needed algebra: $[\sigma^{\mu\nu},\gamma^\rho]=2i(\eta^{\nu\rho}\gamma^\mu-\eta^{\mu\rho}\gamma^\nu)$, which verifies $S^{-1}\gamma^\rho S=\gamma^\rho+\omega^\rho{}_{\nu}\gamma^\nu$ to first order. The <infinitesimal transformation of a Dirac field> is therefore
$$
\boxed{\Psi'(x')=\left(1-\frac i4\omega_{\mu\nu}\sigma^{\mu\nu}\right)\Psi(x)+O(\omega^2).}
$$
At the same coordinate argument, the orbital change must also be included:
$$
\boxed{\delta\Psi(x)=-\omega^\mu{}_{\nu}x^\nu\partial_\mu\Psi(x)-\frac i4\omega_{\mu\nu}\sigma^{\mu\nu}\Psi(x).}
$$
Both formulas describe the same transformation; their arguments differ.
Finally, the gamma adjoint identities imply $\sigma^{\mu\nu\dagger}=\gamma^0\sigma^{\mu\nu}\gamma^0$ and hence the <pseudo-unitarity of the spinor Lorentz representation>, $S^\dagger\gamma^0S=\gamma^0$. It follows that
$$
\bar\Psi'(x')=\bar\Psi(x)S^{-1},\qquad\boxed{\bar\Psi'(x')\Psi'(x')=\bar\Psi(x)\Psi(x).}
$$
Thus the <Dirac scalar bilinear> $\bar\Psi\Psi$ is a <Lorentz scalar>. Infinitesimally the spin terms in $\delta(\bar\Psi\Psi)$ cancel; at fixed coordinates only the ordinary scalar orbital transformation remains.
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