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 (which is not $(0)$ or the field itself). 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 double 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.
There are other properties which fail for non-finitely generated vector spaces in $\sf ZFC$ which are consistent with the failure of choice. The list is long, and these just a few I could write about from the top of my head.
Whether or not any of them is equivalent to the axiom of choice is usually an open (and a difficult) question. But it is usually the case that if a property requires some form of choice (like a basis, or extension of functinoals, etc.) then it can fail in suitable models of $\sf ZF+\lnot AC$.
