Are there axioms intermediate in strength between the axiom of dependent choice and the full axiom of choice?
Is it possible for ZFC to be the inner model of a theory where in the outer model, choice fails?
I ask because I'm wondering if Reinhardt cardinals could exist even if the reals and many large cardinals were well-orderable.
If this is possible, would it have any problematic consequences like the axiom of determinacy is said to?
If not, why not?
The existence of an infinite Dedekind-finite set is consistent with ZF and contradicts both choice (in fact, even countable choice!) and determinacy. AC versus AD isn't a dichotomy by any means, there are lots of other ways things could go. Another more technical example: the existence of an amorphous set contradicts the existence of infinite Dedekind-finite sets + the nonexistence of incomparable Dedekind-finite sets, and both of these principles contradict both AC and AD and are consistent with ZF (okay, in the second case assuming an inaccessible exists; it isn't known if that's necessary yet).
The question of axioms between DC and AC isn't directly related, but yes, there are intermediate axioms: e.g. DC + "AC holds for all sets of size $<\beth_{\omega^2+17}$." (Incidentally, ZF+AD does not prove DC; this is due to Solovay. However, ZF+AD+$V=L(\mathbb{R})$ does imply DC. It is open whether ZF+AD proves DC$(\omega^\omega)$.)
Re: your second paragraph, you're confusing models and theories a bit, but: every model $M$ of ZF has an inner model satisfying ZFC, namely $L^M$. This inner model also satisfies a bunch of other principles, including the generalized continuum hypothesis. This goes the other way too: models of ZFC can have (but don't always have) inner models in which choice fails. For example, the most well-known class of determinacy models is the class of models of the form $L(\mathbb{R})^M$ where $M$ is a model of ZFC + enough large cardinals (if I recall correctly, a proper class of Woodins).
Now you ask specifically about Reinhardts and well-orderings of "small" sets (like $\mathbb{R}$). Well, the usual way to do this would be to argue that forcing $\mathbb{R}$ to be well-orderable can be done with a forcing which is "small" with respect to the Reinhardt and then argue that small forcings preserve Reinhardt-ness. The first is clearly true, but the second I'm not sure about; forcing preservation results often need choice. I suspect that ZF + "There is a Reinhardt" is known to be equiconsistent with ZF + "There is a Reinhardt" + "$\mathbb{R}$ is well-orderable," but I'm not sure.