= Solution
Take the divergence of the <Ampere-Maxwell law> and use <Gauss's law> $\nabla\cdot\boldsymbol E=\rho/\epsilon_0$. Since the divergence of a curl is zero, this gives the <charge continuity equation> $\partial_t\rho+\nabla\cdot\boldsymbol j=0$. Uniform conductivity and <Ohm's law> imply $\nabla\cdot\boldsymbol j=\sigma\nabla\cdot\boldsymbol E=\sigma\rho/\epsilon_0$. Therefore
$$
\boxed{\partial_t\rho=-\frac\sigma{\epsilon_0}\rho,\qquad\rho(t)=\rho_0e^{-t/\tau},\quad\tau=\frac{\epsilon_0}{\sigma}.}
$$
The spatially uniform initial density stays uniform in the interior; $\tau$ is the <charge relaxation> time.
Denote the cylinder radius by $R$. Cylindrical symmetry makes the interior <electric field> radial. <Gauss's law> on a coaxial cylinder of radius $r<R$ and length $\ell$ gives $2\pi r\ell E_r=\rho(t)\pi r^2\ell/\epsilon_0$. Thus
$$
\boxed{\boldsymbol E(r,t)=\frac{\rho_0r}{2\epsilon_0}e^{-t/\tau}\boldsymbol e_r,\qquad\boldsymbol j(r,t)=\frac{\sigma\rho_0r}{2\epsilon_0}e^{-t/\tau}\boldsymbol e_r\quad(r<R).}
$$
The current does not disappear at the interface: the vacuum conducts no current, so charge accumulates on the cylindrical surface. Initially taking the charge to be the stated <volume> charge alone, its <surface charge density> satisfies $\dot\rho_s=j_r(R^-,t)$, giving the <charge relaxation in an isolated conducting cylinder> law
$$
\rho_s(t)=\frac{R\rho_0}{2}(1-e^{-t/\tau}).
$$
The total enclosed charge per unit length is $\pi R^2\rho(t)+2\pi R\rho_s(t)=\pi R^2\rho_0$. Consequently the exterior <electric field> is
$$
\boxed{\boldsymbol E(r,t)=\frac{\rho_0R^2}{2\epsilon_0r}\boldsymbol e_r\quad(r>R),}
$$
independent of time. The boundary jump obeys $E_r(R^+)-E_r(R^-)=\rho_s/\epsilon_0$, as it must. Applying <Gauss's law> to the decaying <volume> charge alone would incorrectly predict a decaying exterior <electric field> and violate <electric charge conservation>. The vector formulas also specify the inward direction if $\rho_0<0$.
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