= Solution
{c}
\b[The quotient.] Evaluation $\mathbb Z[t]\to\mathbb Z[i]$, $t\mapsto i$, is a surjective <ring homomorphism> whose <kernel of a ring homomorphism> is $(t^2+1)$. Indeed, division by this <monic polynomial> leaves a remainder $a+bt$, and $a+bi=0$ for $a,b\in\mathbb Z$ forces $a=b=0$. Taking the further <quotient ring> modulo $p$ gives the explicit <isomorphism>
$$
\boxed{\mathbb F_p[t]/(t^2+1)\xrightarrow{\ \sim\ }\mathbb Z[i]/(p),\qquad \overline t\longmapsto\overline i.}
$$
\b[All fibers, including the generic fiber.] For a point $s\in\operatorname{Spec}\mathbb Z$ with <residue field> $\kappa(s)$, its <scheme-theoretic fiber> is
$$
X_s=\operatorname{Spec}\bigl(\mathbb Z[i]\otimes_{\mathbb Z}\kappa(s)\bigr).
$$
At the <generic point> $s=(0)$, the <residue field> is $\mathbb Q$, and $t^2+1$ is an <irreducible polynomial> over $\mathbb Q$, so $X_{(0)}=\operatorname{Spec}\mathbb Q(i)$. For a <closed point> $(p)$ the <residue field> is $\mathbb F_p$. The permitted criterion for a square root of $-1$ gives three cases. If $p\equiv1\pmod4$, choose $a\in\mathbb F_p$ with $a^2=-1$; the two roots $a,-a$ are distinct, so the <Chinese remainder theorem> gives $\mathbb F_p[t]/(t^2+1)\cong\mathbb F_p\times\mathbb F_p$. If $p\equiv3\pmod4$, the quadratic has no root and is an <irreducible polynomial>, giving the <finite field> $\mathbb F_{p^2}$. Finally, in <characteristic> two, $t^2+1=(t+1)^2$; setting $\epsilon=t+1$ gives the <dual numbers>. Thus
$$
\boxed{X_s\cong\begin{cases}
\operatorname{Spec}\mathbb Q(i),&s=(0),\\
\operatorname{Spec}\mathbb F_p\ \amalg\ \operatorname{Spec}\mathbb F_p,&s=(p),\ p\equiv1\pmod4,\\
\operatorname{Spec}\mathbb F_{p^2},&s=(p),\ p\equiv3\pmod4,\\
\operatorname{Spec}\bigl(\mathbb F_2[\epsilon]/(\epsilon^2)\bigr),&s=(2).
\end{cases}}
$$
The split <scheme-theoretic fiber> has two <closed points>, represented by $(p,i-a)$ and $(p,i+a)$. The inert <scheme-theoretic fiber> has one <closed point>, represented by $(p)$. The <scheme-theoretic fiber> over $(2)$ has one <closed point>, represented by $(1+i)$, but is a <nonreduced scheme>: its nonzero element $\epsilon$ is <nilpotent>. These descriptions retain the <residue fields> and the <nilpotent> structure, which a count of points alone would lose.
\b[The topology.] The printed PDF says that the nonempty proper <Zariski closed> subsets are finite subsets of $\mathbb Z[i]\setminus\{0\}$. Its points are <prime ideals>, so the intended set is $\operatorname{Spec}\mathbb Z[i]\setminus\{(0)\}$.
Put $R=\mathbb Z[i]$. Since $R$ is a <principal ideal domain>, every nonzero <prime ideal> is a <maximal ideal>. Each such <prime ideal> contains a nonzero <Gaussian integer> $z$, and therefore contains the positive integer $z\overline z$. Its contraction to $\mathbb Z$ is consequently a nonzero <prime ideal> $(p)$, so the <scheme-theoretic fibers> just computed list all the <closed points> of $X$. The sole remaining point is the <generic point> $(0)$.
Every proper <Zariski closed> subset is $V((a))$ with $a\ne0$: if its defining <ideal> were zero, it would be all of $X$. A nonzero nonunit $a$ has a finite <unique factorization>
$$
a=u\pi_1^{e_1}\cdots\pi_r^{e_r},\qquad
V((a))=\{(\pi_1),\ldots,(\pi_r)\}.
$$
Here the $\pi_j$ are pairwise nonassociate <prime elements>. Conversely, any finite set of nonzero <prime ideals> $\{(\pi_1),\ldots,(\pi_r)\}$ equals $V((\pi_1\cdots\pi_r))$; the empty set is $V((1))$. Therefore
$$
\boxed{\begin{gathered}
X=\{(0)\}\ \cup\ \{\text{nonzero prime ideals of }\mathbb Z[i]\},\\
\text{proper closed subsets of }X=\text{finite subsets of its closed points}.
\end{gathered}}
$$
In particular, the closure of the <generic point> is $X$, and every nonempty <open subset> contains that <generic point> and all but finitely many <closed points>.
\b[A nonidentity automorphism.] <Complex conjugation>
$$
c:R\longrightarrow R,\qquad a+bi\longmapsto a-bi,
$$
is a <ring automorphism> fixing $\mathbb Z$. It induces the <automorphism> $\sigma=\operatorname{Spec}(c)$ over $\operatorname{Spec}\mathbb Z$. Since $c(i)=-i\ne i$, this is not the identity <morphism of schemes>. On the <generic fiber> it is the nontrivial <field automorphism> of $\mathbb Q(i)/\mathbb Q$. On a split <scheme-theoretic fiber> it swaps the two <closed points>, since $i-a$ maps to $-(i+a)$. On an inert <scheme-theoretic fiber> it fixes the only point but induces the nonidentity <finite-field Frobenius automorphism> $z\mapsto z^p$ on its <residue field>, because $i^p=-i$. On the <scheme-theoretic fiber> over $(2)$ it is the identity on the entire <scheme>: $-i=i$ in $R/(2)$, so $\epsilon=i+1$ is fixed. Thus
$$
\boxed{\sigma:\quad\text{generic conjugation; split points exchanged; inert Frobenius; identity over }2.}
$$
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