I have the following exercise from Dirk Van Dalen's Logic and Structure:
Here with language of identity we mean the language with no extralogical symbols (i.e. no symbols of predicates, functions or constants), the $=$ being a logical symbol. Formulas $I_1, I_2, I_3$ are the axioms for the equivalence relation, i.e. respectively
$$ I_1 := \forall x (x=x)\\ I_2 := \forall xy (x=y \rightarrow y=x) \\ I_3 := \forall xyz(x=y \land y=z \rightarrow x=z) $$
My doubt is: being the language so simple, an isomorphism is just a bijection, so how it is possible that the theory is not $\aleph_0$-categorical? Aren't two countable sets necessarily in bijection?
Hint: For the countable case, think along these lines: Is every element in your domain named by one of the constants? What happens if they are not?
Edit1: Further to your comment; suppose that every element of $M\models{T}$ is named by one of the constants and $N\models{T}$ has exactly one element that is not named by the constants (here $M$, $N$ are countable). Now consider any bijection $f$ between them. Now use the definition of $L$ isomorphism to show that $f$ is not an L-isomorphism (You say that you think any bijection is an L-isomorphism. This is not true).