Well they ask me:
Be $\mathbb{A}$ and $\mathbb{B}$ $L$-structures and $A \subset B$ where $A \text{ and } B$ are the universe of $\mathbb{A} \text{ and } \mathbb{B}$. Give an example of $\mathbb{A} \text{ and } \mathbb{B}$ where $\mathbb{A}$ is not a substructure of $\mathbb{B}$.
Well I start thinking about this and the problem of $\mathbb{A}$ being a substructure of $\mathbb{B}$ must be on the constants or in the functions. But I don't know how to do it, because all of my ideas are wrong in the point that I define something that fulfill $\mathbb{A}$ is not a substructure of $\mathbb{B}$ but also $\mathbb{A}$ is not a $L$-structure, like in this case:
$L = \{ f, 0 \}$ with $f(x) = x+1$ and $\mathbb{A} = \lbrace \text{even numbers} , f^A,0^A \rbrace$ and $\mathbb{B} = \lbrace \mathbb{R} , f^B,0^B \rbrace$ as I said $\mathbb{A}$ is not a L-structure.
I think you do not need to constrain by that the same symbol must have the same definition, and the following example works:
Let $A=B=\omega$. Consider the following orders over $\omega$: $\le_0$ is the usual order, and $\le_1$ is an order that enumerates natural numbers as follows: $$\cdots 4 <_1 2 <_1 0 <_1 1 <_1 3 <_1 5 <_1 \cdots$$ (that is, $(\omega,<_1)$ is isomorphic to $(\mathbb{Z},<)$.) Then both $(\omega,<_0)$ and $(\omega,<_1)$ shares the same domain, but the former is not a substructure of the latter.