= Solution
{c}
\b[No.] Let $k=\mathbb F_2$ be the <field with two elements>, put $A=k[\varepsilon]/(\varepsilon^2)$, and take the <closed immersion>
$$
\boxed{f:\operatorname{Spec}k\hookrightarrow\operatorname{Spec}A}
$$
induced by the <ring homomorphism> $q:A\twoheadrightarrow k$ with $q(\varepsilon)=0$.
For every <field> $K$, a <ring homomorphism> $\alpha:A\to K$ must send $\varepsilon$ to zero: $\alpha(\varepsilon)^2=0$, and a <field> has no nonzero <nilpotent elements>. Consequently $\alpha$ factors uniquely through $q$. Passing from <ring homomorphisms> to <morphisms of schemes> between <affine schemes> gives a <bijection>
$$
f(\operatorname{Spec}K):\operatorname{Hom}_{\mathrm{Sch}}(\operatorname{Spec}K,\operatorname{Spec}k)
\xrightarrow{\ \sim\ }
\operatorname{Hom}_{\mathrm{Sch}}(\operatorname{Spec}K,\operatorname{Spec}A).
$$
This concerns all <morphisms of schemes>, without requiring them to be over $k$: if $K$ does not have <characteristic> two, both sets are empty; otherwise each contains exactly one element.
Nevertheless, $f$ is not an <isomorphism of schemes>. Its map on <global sections> is $q$, whose <kernel of a ring homomorphism> contains the nonzero element $\varepsilon$, so $q$ is not a <ring isomorphism>. The target is the <nonreduced scheme> defined by the <dual numbers>, whereas the source is a <reduced scheme>. This illustrates why <field-valued points do not detect nilpotents>.
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