2 retraction

This is wrong. But see my other answer arguing the opposite direction.

What's wrong is that Bertini's theorem fails over $\mathbb Z$. It works over infinite fields in single pencils. A version works over finite fields by allowing arbitrary degree and thus infinitely many choices. But high degree over $\mathbb Z$ is bad for smoothness. As Will points out, even in $P^2_{\mathbb Z}$, a high degree hypersurface is not smooth.

I think that Godeaux-Serre varieties exist integrally. Choose a prime $p$, let $G$ be the cyclic group of order $p$ and let $Z[G]$ be the group ring. Then the projectivization of the $n$-th power of the ring group $P(Z[G]^n)$ is an $pn-1$-dimensional variety with an action of $G$, generically free, with fixed set a disjoint union of $P^{n-1}$; 1 copy at $p$ and $p$ copies away from $p$. The quotient is not smooth, but a generic complete intersection of codimension $n$ misses the singular set and thus is smooth with fundamental group $G$.

I have never seen Godeaux-Serre varieties used in same characteristic, but when I looked up Igusa's example, I saw it asserted that not only does the construction work, they have non-reduced Picard scheme, evading Will's attack. But does this generic complete intersection argument work globally?

I think that Godeaux-Serre varieties exist integrally. Choose a prime $p$, let $G$ be the cyclic group of order $p$ and let $Z[G]$ be the group ring. Then the projectivization of the $n$-th power of the ring group $P(Z[G]^n)$ is an $pn-1$-dimensional variety with an action of $G$, generically free, with fixed set a disjoint union of $P^{n-1}$; 1 copy at $p$ and $p$ copies away from $p$. The quotient is not smooth, but a generic complete intersection of codimension $n$ misses the singular set and thus is smooth with fundamental group $G$.