Converse to a proposition regarding minimal structures

Question

Suppose $M$ is an infinite structure where the only parameter-free definable subsets is either finite or cofinite. Must $M$ be a minimal structure, that is one where all the definable subsets with parameters are finite or cofinite? And what if we strengthen the condition to say that every parameter-free definable set is either empty or the whole set $M$?

Answer 1

No. Your stronger condition, that every parameter-free definable subset of $M$ either empty or the whole set $M$, is equivalent to the statement that there is exactly one complete $1$-type over the empty set modulo $\text{Th}(M)$.

A sufficient condition for this is that $M$ (or any structure elementarily equivalent to $M$!) has a transitive automorphism group, i.e. for all $a,b\in M$, there exists $\sigma\in \text{Aut}(M)$ such that $\sigma(a) = b$.

There are lots of structures with transitive automorphism groups which are not minimal. For example, $(\mathbb{Q};<)$, or $(\mathbb{Z};<)$, or the random graph, or $(\mathbb{N};E)$, where $E$ is any non-trivial equivalence relation for which every equivalence class is infinite (for example the relation "equivalent mod $n$" with $n\geq 2$).