Exercise goes as follows:
Assume $T$ is $\mathcal{L}$-theory, where $|\mathcal{L}|$ is finite and it has a model $\mathcal{M}$ which is isomorphic to it's own proper substructure. Show that $T$ has a countable model isomorphic to it's own proper substructure.
To begin with, I'm confused of how it is possible that some model $\mathcal{M}$ has a proper substructure $\mathcal{N}$ such that there is isomorphic $\mathcal{L}$-embedding $\eta: N \to M$. Since $\eta$ is isomorphic, how it is possible that $\mathcal{N}$ is proper substructure? I think proper would mean $N \subsetneq M$.
I would appreciate any help of explaining it.
EDIT: Now I understand the text of the exercise fully, but how can I show it is true for any such $T$? I would appreciate any hints on how can I prove this.
EDIT2: So here is my attempt to prove it:
Let $T$ be a theory in a finite language $\mathcal{L}$ and let $\mathcal{L}$-structures $\mathcal{M}$, $\mathcal{N}$ be such that $\mathcal{M} \models T$, $\mathcal{N} \subsetneq \mathcal{M}$ and $\mathcal{N} \cong \mathcal{M}$. Let $\mathcal{L}^* = \mathcal{L}\cup \{f\}$, where $f$ a new unary function symbol and let $\varphi = \varphi_1 \land \varphi_2 \land \varphi_3$, where $$ \varphi_1 \text{ is } \forall v_1 \forall v_2 (f(v_1) = f(v_2) \Rightarrow v_1 = v_2) \quad \text{(}f\text{ is an injection)} $$ $$ \varphi_2 \text{ is } \forall v_2 \exists v_1 (f(v_1) = v_2) \quad \text{(}f\text{ is onto)} $$ $$ \varphi_3 \text{ is } \exists v_3 \forall v_1 \forall v_2 (f(v_1) = v_2 \Rightarrow \neg(v_1 = v_3)) \quad \text{(image of }f\text{ is not equal to its domain)} $$ and $$ T^* = T \cup \{ \varphi \} $$ Let $\eta: \mathcal{M} \to \mathcal{N}$ be an $\mathcal{L}$-isomorphism. If we let $f^{\mathcal{M}}(a) = \eta(a)$, then clearly $\mathcal{M} \models T^*$ as an $\mathcal{L}^*$-structure. Now, by Lowenheim–Skolem theorem, there exist countable elementary $\mathcal{L}^*$-substructure $\mathcal{S} \preceq \mathcal{M}$. Therefore, by the construction of $T^*$, $f^{\mathcal{S}}: S \to f^{\mathcal{S}}[S]$ is an isomorphism. Moreover, by $\varphi_3$, $f^{\mathcal{S}}[S] \neq S$, so $f^{\mathcal{S}}$ defines a proper isomorphic $\mathcal{L}^*$-substructure of $\mathcal{S}$. Finally, since $\mathcal{S} \models T^*$, then $\mathcal{S} \models T$, so $\mathcal{S}$ is a countable model of $T$ isomorphic to its own proper substructure.
Hint: Add an extra unary function symbol $f$ and choose an $L\cup \{f\}$ theory whose models (as $L$-structures) are exactly the $M\models T$ such $f$ is an isomorphism between $M$ and $f[M]\neq M$, and notice that by hypothesis, this theory is consistent. Then apply Lowenheim-Skolem.