$\mathbb R$ is an uncountably dimensional vector space over $\mathbb Q.$ We can define as many endomorphisms of this vector space as we want by picking their values on the elements of the basis. However, to have a basis we need to use the axiom of choice, so this way is non-constructive. Any $\mathbb R-$ endomorphism of $\mathbb R$ is also a $\mathbb Q-$endomorphism. But can we give a concrete example of a $\mathbb Q-$endomorphism of $\mathbb R$ that is not an $\mathbb R-$ endomorphism?
Answer
Using the axiom of choice this is trivial. Choose a Hamel basis, and take any permutation of it.
However without the axiom of choice it is perfectly feasible to have a model in which there is no Hamel basis, that is $\mathbb R$ as a vector space over $\mathbb Q$ has no basis.
Of course that one does not need a basis to have endomorphisms, however we can make sure that indeed there are no endomorphisms of $\mathbb R$ over $\mathbb Q$. From such automorphism we can generate non-measurable sets, so if we happen to live in a model of ZF in which every set of real numbers is Lebesgue measurable there can be no such endomorphism.
Natural examples for such models are models of ZF when assuming The Axiom of Determinacy instead of The Axiom of Choice; or Solovay's model.
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