To add the proof for my claim in Todd's answer, which essentially repeats Läuchli's original [1] arguments with minor modifications (and the addition that the resulted model satisfies $DC_\kappa$).
We will show that it is consistent to have a model in which $DC_\kappa$ holds, and there is a vector space over $\mathbb F_2$ which has no linear functionals.
Assume that $M$ is a model of $ZFA+AC$ and that $A$, the set of atoms has $\lambda>\kappa$ many atoms, where $\lambda$ is a regular cardinal. Endow $A$ with a structure of a vector space over $\mathbb F=\mathbb F_2$. Now consider the permutation model $\frak M$ defined by the group of linear permutations of $A$, and by ideal of supports generated by subsets of dimension $\le\kappa$.
Denote by $\operatorname{fix}(X)$ the permutations which fix every element of $X$, by $\operatorname{sym}(X)$ the permutations that fix $X$ as a set, and by $[E]$ the span of $E$ as a subset of $A$. We say that $E\subseteq A$ is a support of $X$ if $\pi\in\operatorname{fix}(E)\Rightarrow\pi\in\operatorname{sym}(X)$.
Final word of terminology, since $A$ will play both the role of set of atoms as well the vector space, given $U\subseteq A$ the complement will always denote a set complement, whereas the direct complement will be used to refer to a linear subspace which acts as a direct summand with $U$ in a decomposition of $A$.
Claim 1: If $E$ is a subset of $A$ then $\operatorname{fix}(E)$ is the same as $\operatorname{fix}([E])$.
Proof: This is obvious since all the permutations considered are linear. $\square$
From this we can identify $E$ with its span, and since (in $M$) the $[E]$ has the same cardinality of $E$ we can conclude that without loss of generality supports are subspaces.
Claim 2: $\frak M$$\models DC_\kappa$.
Proof: Let $X$ be some nonempty set, and $\lt$ a binary relation on $X$, both in $\frak M$. In $M$ we can find a function $f\colon\kappa\to X$ which witness $DC_\kappa$ in $V$.
Since $\frak M$ is transitive, we have that $\alpha,f(\alpha)\in\frak M$ and thus $\langle\alpha,f(\alpha)\rangle\in\frak M$. Let $E_\alpha$ be a support for $\lbrace\langle\alpha,f(\alpha)\rangle\rbrace$ then $\bigcup_{\alpha<\kappa} E_\alpha$ is a set of cardinality $<\kappa^+$ and thus in our ideal of suports. It is simple to verify that this is a support of $f$, therefore $f\in\frak M$ as wanted. $\square$
Claim 3: If $x,y\in A$ are nonzero (with respect to the vector space) then in $M$ there is a linear permutation $\pi$ such that $\pi x=y$ and $\pi y=x$.
Proof: Since $x\neq y$ we have that they are linearly independent over $\mathbb F$. Since we have choice in $M$ we can extend this to a basis of $A$, and take a permutation of this basis which only switches $x$ and $y$. This permutation extends uniquely to our $\pi$.
Claim 4: If $U\subseteq A$ and $U\in\frak M$ then either $U$ is a subset of a linear subspace of dimension at most $\kappa$, or a subset of the complement of such space.
Proof: Let $E$ be a support of $U$, then every linear automorphism of $A$ which fixes $E$ preserves $U$. If $U\subseteq [E]$ then we are done, otherwise let $u\in U\setminus [E]$ and $v\in A\setminus [E]$, we can define (in $M$ where choice exists) a linear permutation $\pi$ which fixes $E$ and switches $u$ with $v$. By that we have that $\pi(U)=U$ therefore $v\in U$, and so $U=A\setminus[E]$ as wanted. $\square$
Claim 5: If $U\subseteq A$ is a linear proper subspace and $U\in\frak M$ then its dimension is at most $\kappa$.
Proof: Suppose that $U$ is a subspace of $A$ and every linearly independent subset of $U$ of cardinality $\le\kappa$ does not span $U$, we will show $A=U$. By the previous claim we have that $U$ is the complement of some "small" $[E]$.
Now let $v\in A$ and $u\in U$ both nonzero vectors. If $u+v\in U$ then $v\in U$. If $u+v\in [E]$ then $v\in U$ since otherwise $u=u+v+v\in[E]$. Therefore $v\in U$ and so $A\subseteq U$, and thus $A=U$ as wanted.$\square$
Claim 6: If $\varphi\colon A\to\mathbb F$ a linear functional then $\varphi = 0$.
Proof: Suppose not, for some $u\in A$ we have $\varphi(u)=1$, then $\varphi$ has a kernel which is of co-dimension $1$, that is a proper linear subspace and $A=\ker\varphi\oplus\lbrace 0,u\rbrace$. However by the previous claim we have that $\ker\varphi$ has dimension $\kappa$ at most, and without the axiom of choice $\kappa+1=\kappa$, thus deriving contradiction to the fact that $A$ is not spanned by $\kappa$ many vectors.
Aftermath: There was indeed some trouble in my original proof, after some extensive work in the past two days I came to a very similar idea. However with the very generous help of Theo Buehler which helped me find the original paper and translate parts, I studied Läuchli's original proof and concluded his arguments are sleek and nicer than mine.
While this cannot be transferred to $ZF$ using the Jech-Sochor embedding theorem (since $DC_\kappa$ is not a bounded statement), I am not sure that Pincus' transfer theorem won't work, or how difficult a straightforward forcing argument would be.
Lastly, the original Läuchli model is where $\lambda=\aleph_0$ and he goes on to prove that there are no non-scalar endomorphisms. In the case where we use $\mathbb F=\mathbb F_2$ and $\lambda=\aleph_0$ we have that this vector space is indeed amorphous which in turn implies that very little choice is in such universe.
Bibliography:
- Läuchli, H. Auswahlaxiom in der Algebra. Commentarii Mathematici Helvetici, vol 37, pp. 1-19.
Best Answer
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 $\sf ZF+\lnot AC$.