= Solution
Use the <first-order language> with one <equivalence relation> $E$. Let $\mathcal A$ have exactly two <equivalence classes>, one countably infinite and one uncountable, with no further <first-order structure>. Its <complete theory> says that there are exactly two classes and both are infinite.
This <first-order theory> has <quantifier elimination>. To eliminate one existential <logical variable> from a <conjunction> of <atomic formulas> and negated <atomic formulas>, a <logical equality> either fixes its value, or finitely many inequalities exclude finitely many points. Its required equivalence-class membership is determined by the finitely many existing classes represented among the parameters. A required unused class exists exactly when fewer than two classes have already been used. Once its class is possible, infinitude supplies a point avoiding the finite exclusions. These tests are quantifier-free <logical equality> and equivalence conditions on the parameters. <Disjunctive normal form> and <induction> eliminate all <quantifiers> and also show completeness.
Thus a finite <tuple>'s <complete type> is isolated by its finite equality-and-equivalence pattern. Every realized finite-tuple <partial type> is principal, so $\mathcal A$ is atomic. In particular all individual elements have the same <complete one-type>, isolated by $x=x$.
Choose $a$ in the <countable> class and $b$ in the uncountable class. Any automorphism taking $a$ to $b$ would give a <bijection> from their whole <equivalence classes>, impossible because their cardinalities differ. Therefore
$$
\boxed{\mathcal A\text{ is atomic but not homogeneous}.}
$$
Its uncountability is precisely what avoids the conclusion of (iii).
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