Some things about vector spaces which are *consistent* with the failure of choice:

1. 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 http://mathoverflow.net/questions/93242/sizes-of-bases-of-vector-spaces-without-the-axiom-of-choice for more details.

2. 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 http://mathoverflow.net/questions/80765/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.

3. 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: http://mathoverflow.net/questions/49388/is-the-non-triviality-of-the-algebraic-dual-of-an-infinite-dimensional-vector-sp/79437#79437

4. 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 http://mathoverflow.net/questions/49351/does-the-fact-that-this-vector-space-is-not-isomorphic-to-its-double-dual-requir/93201#93201 for details.

5. 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 http://mathoverflow.net/questions/64219/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.