It is well-known that every ordered field contains a subfield that is isomorphic to $\langle \Bbb Q,<,+,\cdot,0,1 \rangle$. Let $\langle A,<,+,\cdot,0',1' \rangle$ be an ordered field. I would like to ask two questions.
Is it possible for $\langle A,<,+,\cdot,0',1' \rangle$ to contain two different subfields that are both isomorphic to $\langle \Bbb Q,<,+,\cdot,0,1 \rangle$?
Is it possible for $\langle \Bbb Q,<,+,\cdot,0,1 \rangle$ to be isomorphic to a proper subfield of itself? (proper means the underlying set of the subfield is a proper subset of $\Bbb Q$)
Thank you for your help!
No. Each element of $\mathbb Q$ satisfies an equation of the form $$ x = \frac{1+1+\cdots+1+1}{1+1+\cdots+1+1} \qquad\text{or}\qquad x = -\frac{1+1+\cdots+1+1}{1+1+\cdots+1+1} $$ Since the right-hand side is an equation in the language of field theory (which can be evaluated in every field of characteristic 0), for each rational number there is at most one element of every field that it can map to by isomorphism.
For your (1) this means that there is at most one isomorphism $\mathbb Q\to A$, so if such an isomorphism exists its range is unique.
The impossiblity of (2) follows from (1) by setting $A=\mathbb Q$.