Can the set of endomorphisms of $(\mathbb{R},+)$ have cardinality strictly between $\mathfrak c^ c$ and $\frak c$?

Let $$\mathfrak{c}$$ be the cardinality of the continuum. It is well known that in ZFC that the reals under addition has $$\frak{ c^c}$$(cardinality of maps to itself, and in ZFC this is the same as $$2^\frak c$$) self homomorphisms which you can find by thinking of the reals as a vector space over the rationals and permuting basis elements(which requires some amount if choice)

It is also well known, at least in set theory circles, that it is consistent with ZF that there are only continuous homomorphisms and a little argument shows that there are $$\mathfrak{c}$$ of those.

Adding axiom of choice is a lot, so maybe you don't have to add so many new homomorphisms.

Question(s): Can the set of self homomorphisms of the reals have cardinality strictly between the two cardinalities discussed above?(see edit below but the more natural upper bound is $$\frak {c^c}$$) With cardinality gaps can you chose which cardinal in between arbitrarily? I would also be interested in the same question except isomorphisms instead of endomorphisms(where the more natural upper bound is $$\frak{c}!$$).

(This is inspired by a question I recently answered, and technically an answer to this could be an answer to that, although this is more focused)

Edit: There might be some subtlety in the upper bound cardinal, since without choice things get a weird where $$\kappa^\kappa$$, $$\kappa !$$(cardinality of bijections), and $$2^\kappa$$ might be different. Not sure if that is true for $$\frak c$$ but we at leat have the set of endomorphisms has cardinality less than or equal to $$\frak {c^c}$$. A paper by Dawson and Howard, Factorials of infinite cardinals, says that $$\kappa! < 2^\kappa$$, $$2^\kappa < k!$$, or the cardinals are incomparable are all possible.

I have asked this question on mathoverflow.

• @BrianMoehring: I don't think this depends on whether GCH holds at the metalevel. Note that when we ask "can such-and-such be true" in set theory, this is always jargon for "is there a model of set theory that satisfies such-and-such", or equivalently, "is the theory ZF+such-and-such consistent". – hmakholm left over Monica Sep 24 '19 at 2:53
• (In fact, I'm pretty sure that it doesn't depend on whether GCH holds at the metalevel, since consistency questions are arithmetical, and you can both make GCH hold and make GCH fail without modifying the naturals). – hmakholm left over Monica Sep 24 '19 at 2:56
• @BrianMoehring Remember we're working without choice here - GCH is jumping the gun a bit. – Noah Schweber Sep 24 '19 at 3:01
• Definitely not "every cardinal" is consistent. For example, if $\aleph_1\nleq\frak c$, then $\aleph_1\times\frak c<c^c$, but there is no group structure on that set, since a group structure implies comparability (as per the usual proof that group structure on every non-empty set implies the axiom of choice). – Asaf Karagila Sep 24 '19 at 7:22
• @LeeMosher It follows from a couple things. First there is the Solovay model which provides a model of ZF where ever set of reals is Baire measurable(this is the hard set theory part). Then there is a theorem of Pettis, which "isn't hard" and can be found in Automatic continuity of group homomorphisms by Christian Rosendal Thm 2.2, which says that any Baire measurable homomorphism between Polish groups is actually continuous(nothing about ZF or ZFC). Putting this together you get that there are actually only continuous homomorphisms. – Paul Plummer Sep 24 '19 at 16:50