Let $(\mathbb{Q},0,1,<)$ be the rational numbers, considered as structures in the language $\mathcal{L}=\{0,1,<\}$.
Question is that:
Explicitly define an isomorphism $i:(\mathbb{Q},0,1,<) \rightarrow (\mathbb{Q},0,1,<)$ other than the identity map $i(q)=q$.
My Answers are $i(q) = q + 1$ or $i(q) = q + q$ or $i(q) = q \cdot q$.
Are these answers okay for this question ?
Also, I want to ask does the first-order Language include the function symbol $+$ and the function symbol $\cdot$ ?
None of these work. The isomorphism needs to take $0\mapsto 0$ and $1\mapsto 1,$ so that rules out the first two. It also needs to preserve order, which rules out the third. Instead, think about an order preserving map that fixes $0$ and $1.$ For instance, anything of the form $ax+b$ where $a,b\in \mathbb Q$ and $a>0$ is an order preserving map $\mathbb Q\to \mathbb Q$ (i.e. an isomorphism of $(\mathbb Q,< )).$ It will take some thinking to figure out how to use this idea to get something that also fixes $0$ and $1,$ since the only linear function that will preserve $0$ and $1$ is $a=1,b=0,$ which is just the identity... you need to play around with related ideas.
The language you have written down doesn't include $+$ or $\cdot$... they would be written down there if they were included. This just means our ismorphism doesn't need to preserve these functions. There is nothing wrong with using these functions to define our isomorphism, which is something we do in the metatheory, not the language itself.
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Here is an explicit isomorphism: Let $$i(q) = \begin{cases} q& q<1\\2q-1&q\ge 1\end{cases}.$$ It's easy to verify the $i$ is a bijction $\mathbb Q\to \mathbb Q$ with $i(0)=0,$ $i(1)=1,$ and $i(x)<i(y)\iff x<y,$ which is what it means to be an isomorphism.
You had a good idea of trying $i(q)=q^3,$ which has $i(0)=0,$ $i(1)=1,$ and $i(x)<i(y)\iff x<y,$ however, it is not an isomorphism since it is not a bijection. Its image only consists of rationals that are perfect cubes, not all of the rationals, so it is not surjective. It would definitely be a nontrivial isomorphism if the structure were $(\mathbb R,0,1,<)$ rather than $(\mathbb Q,0,1,<).$