$\mathbb R$ 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?
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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.