Some things about vector spaces which are consistent with the failure of choice:
Vector spaces may have bases of different cardinality. In particular, this means that the notion of "dimension" is not well-defined. It follows from the Boolean Prime Ideal theorem (which is strictly weaker than $\sf AC$ itself) that if there is a basis, then its cardinality is unique. See Sizes of bases of vector spaces without the axiom of choice for more details.
The existence of a basis is no longer hereditary. That is, it is consistent that there is a vector space which has a basis, but it has a subspace which doesn't have a basis. You can find the example in Goldstern's answer If $V$ is a vector space with a basis. $W\subseteq V$ has to have a basis too?, and what is even more interesting is the fact that the vector space without a basis has a direct complement which has a well-ordered basis.
It is consistent that there is a vector space, that all its endomorphisms are scalar multiplications. In particular every non-zero endomorphism is an automorphism, and this answers yours final question. Indeed every non-zero endomorphism is an injective endomorphism and an automorphism. These spaces were the main topic of my masters thesis, where I somewhat extended Lauchli's original result (and construction) of such spaces. You can find somewhat of an outline of the general result here: Is the non-triviality of the algebraic dual of an infinite-dimensional vector space equivalent to the axiom of choice?
It is consistent that there is a vector space, which is not finitely generated, which is (naturally) isomorphic to its algebraic dual. In particular this can be $\ell_2$. See my answer at Does the fact that this vector space is not isomorphic to its double-dual require choice? for details.
It is consistent that there is a vector space which is not the direct sum of any two of its proper subspaces. Every vector space with this property has also the property that every endomorphism is a scalar multiplication, but the inverse is not true. I used to think that the existence of such indecomposable vector space is provable from the negation of the axiom of choice, but now I don't think so anymore (because I realized that I have no clarity on this topic to have an intuition based guess). See Indecomposable vector spaces and the axiom of choice for some details.
There are other properties which fail, or become consistently true for non-finitely generated vector spaces. The list is long, and these just a few I could write about from the top of my head.