= Solution
Let $\mathbf n=\mathbf r/a$ be the <unit normal> from the <electrical conductor> into the <vacuum>. The <electrostatic boundary conditions at a conductor>, together with the vanishing interior <electric field>, give
$$
\boxed{\mathbf E_{\rm out}=\frac{\sigma}{\epsilon_0}\mathbf n,\qquad
\sigma=\epsilon_0\mathbf E_{\rm out}\cdot\mathbf n=-\epsilon_0\left.\partial_r\phi\right|_{a^+}.}
$$
Write $u=\cos\theta$ in <spherical coordinates>, $D=|\mathbf r-\mathbf r_0|$ on $r=a$, and $z^\star=a^2/r_0$. Differentiating the <electric potential> at fixed angles gives
$$
\left.\partial_r\phi\right|_{a^+}
=\frac q{4\pi\epsilon_0D^3}\left[-(a-r_0u)+\frac{r_0^2}{a^2}\left(a-\frac{a^2}{r_0}u\right)\right]
=\frac q{4\pi\epsilon_0D^3}\left(\frac{r_0^2}{a}-a\right).
$$
Here the distance identity in the preceding answer was used before simplifying. Hence the induced <surface charge density> is
$$
\boxed{\sigma(\mathbf r)=\frac{q(a-r_0^2/a)}{4\pi|\mathbf r-\mathbf r_0|^3},\qquad |\mathbf r|=a.}
$$
For positive $q$ this <surface charge density> is negative everywhere, with largest magnitude nearest the <point charge>.
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