Let $k$ be a field and $\mathbf{P}$ a projective space over $k$. If we accept the axiom of choice (AC), then $\mathbf{P}$ has a basis and a dimension $m$, and if $m$ is finite, the automorphism group is given by $\mathbf{P\Gamma L}_{m + 1}(k)$, a semidirect product of $\mathbf{PGL}_{m + 1}(k)$ and the automorphism group of $k$. (I take it that this is also true in infinite dimensions.)
On the other hand, if we work in ZF without AC, then there are models in which there exist projective spaces without a basis, without a well-defined dimension, and in this post it is even shown that there exist projective spaces with a trivial automorphism group.
Also, in ZF without AC, it is consistent to say that there are projective spaces of infinite dimension in which any subspace has a finite dimension. So for instance, there are no hyperplanes.
My question might be very stupid, but taken these weird projective spaces into account, is it clear in ZF without AC, whether all automorphisms of a projective space are (semi)linear, that is, sitting in $\mathbf{P\Gamma L}_{\{\cdot\}}(k)$ ? I did not fill out the dimension, as there might not be a dimension; the latter group is defined as the general linear group coming from the underlying vector space, together with the field automorphisms like before.
(Even when I write this, I seem to get in trouble because I don't know how to readily define the action of an automorphism of $k$ on a point of the space, since we don't necessarily have coordinates ...)
