 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. ThereforeIts uncountability is precisely what avoids the conclusion of (iii).