Here is a little progress towards AC.
Theorem.
ICF implies the dual Cantor-Schröder-Bernstein
theorem, that is $X$ surjects onto $Y$ and $Y$ surjects onto $X$,
then they are bijective.
Proof. You explain this in the edit to the question. If
$X\twoheadrightarrow Y$, then $2^Y\leq 2^X$ by taking pre-images,
and so if also $Y\twoheadrightarrow X$, then $2^X\leq 2^Y$ and so
$X\sim Y$ by ICF. QED
Theorem. ICF implies that there are no infinite D-finite
sets.
Proof. (This is a solution to the exercise that you mention.) If $A$ is infinite and
Dedekind-finite, then let $B$ be the set of all finite
non-repeating finite sequences from $A$. This is also D-finite,
since a countably infinite subset of $B$ easily gives rise to a
countably infinite subset of $A$. But meanwhile, $B$ surjects onto
$B+1$, since we can map the empty sequence to the new point, and
apply the shift map to chop off the first element of any sequence.
So $B$ and $B+1$ surject onto each other, and so by the dual
Cantor-Schöder-Bernstein result, they are bijective,
contradicting the fact that $B$ is D-finite. QED
Here is the new part:
Theorem. ICF implies that $\kappa^+$ injects into
$2^\kappa$ for every ordinal $\kappa$.
Proof. We may assume $\kappa$ is infinite. Notice that
$2^{\kappa^2}$ surjects onto $\kappa^+$, since every
$\alpha<\kappa$ is coded by a relation on $\kappa$. Since
$\kappa^2\sim\kappa$, this means
$2^\kappa\twoheadrightarrow\kappa^+$ and consequently
$2^{\kappa^+}\leq 2^{2^\kappa}$, by taking pre-images. It follows that
$2^{2^\kappa}=2^{2^\kappa}\cdot 2^{\kappa^+}=2^{2^\kappa+\kappa^+}$ and
so by ICF we get $2^\kappa+\kappa^+=2^\kappa$, which implies
$\kappa^+\leq 2^\kappa$, as desired. QED
This conclusion already contradicts AD, for example, since AD
implies that there is no $\omega_1$ sequence of distinct reals,
which violates the conclusion when $\kappa=\omega$. In particular,
this shows that ICF implies $\neg$AD, and so in every AD model, there are sets of different cardinalities, whose power sets are equinumerous.
Let me add a few examples:
(1) If we start with a supercompact cardinal $\kappa$, and force with $Add(\omega, \kappa)$, then in the extension the cardinal $\kappa=2^\omega$ becomes generically supercompact. The same holds for many other large cardinals.
(2) If we start with a weakly compact cardinal, we can find a generic extension in which $2^\omega=\kappa$ is (the least) weakly Mahlo, and tree property holds at $\kappa.$ This result is due to Boos ``Boolean extensions which efface the Mahlo property''.
(3) The consistency of the theory $ZFC$+ "there is a supercompact cardinal'' implies the consistency of the theory $ZFC$ + "there exist a uniform measure $μ$ on the cardinal $2^ω$ and a set $X⊆2^ω$ of positive $μ$-measure such that for every $y∈X$ there is a uniform measure on $y$ which is $|y|$-additive.''
(4) The consistency of the theory $ZFC$+ "there is a measure concentrating on compact cardinals'' implies the consistency of the theory $ZFC$ + "there exist a uniform measure $μ$ on the cardinal $2^ω$ and a set $X⊆2^ω$ of positive $μ$-measure such that for every $y∈X$ there is a uniform measure on $2^ω$ which is $|y|$-additive.''
For (3) and (4) see Some combinatorial properties of measures
(5) Under $PFA, 2^\omega=\aleph_2$ has some large cardinal properties.
Best Answer
Yes, you are right. This is a theorem of Specker. If there are no intermediate cardinals between $A,\mathcal P(A)$ and $\mathcal{P(P(}A))$, then $A$ can be well-ordered.
You can find nice details in:
Link to the paper on Kanamori's website.
And on this relevant thread.
I had discussed this answer with Yair Hayut earlier this evening, and then I noticed the following peculiar thing.
Suppose that $\sf CH$ holds for $\Bbb R$ and $\aleph_1$, then $2^{\aleph_0}=\aleph_1$ or $2^{\aleph_0}=\aleph_2$. In either case, it is well-orderable.
First we observe that since there is a definable surjection from $\Bbb R$ onto $\omega_1$, it is necessarily the case that $\aleph_1$ and $2^{\aleph_1}$ are both $\leq2^{2^{\aleph_0}}$, with $\aleph_1$ being strictly smaller. Therefore $\aleph_1\leq2^{\aleph_0}\leq2^{\aleph_1}\leq2^{2^{\aleph_0}}$, but our assumptions tell us it is impossible that all three are strict inequalities. Either the middle one is equality, or both the left and right ones are equality (but not both, of course).
If $2^{\aleph_0}<2^{\aleph_1}$, then we have that $\aleph_1\leq2^{\aleph_0}<2^{\aleph_1}$, and therefore $2^{\aleph_0}=\aleph_1$.
If $2^{\aleph_1}=2^{\aleph_0}$ then $\aleph_2<2^{2^{\aleph_0}}$ (since there is a definable surjection from $2^{\omega_1}$ onto $\omega_2$, so by this assumption there is a surjection from $\Bbb R$ onto $\omega_2$). If $2^{\aleph_0}$ and $\aleph_2$ are incomparable then we get $2^{\aleph_0}<2^{\aleph_0}+\aleph_2<2^{2^{\aleph_0}}$ which is a contradiction (this is the same argument by $\aleph_1\leq2^{\aleph_0}$). Therefore $\aleph_2\leq 2^{\aleph_0}$. If they are equal, wonderful.
Otherwise, $\aleph_1<\aleph_2<2^{\aleph_0}=2^{\aleph_1}$ which is a contradiction to our assumption that $\sf CH(\aleph_1)$ holds.
So while it doesn't decrease the number of sets which required here, it does give a nice twist on the things. I'm not sure if we can get rid of the assumption of $\sf CH(\aleph_1)$, maybe at the cost of requiring $\sf CH(\aleph_2)$, and to remove that assumption we would need to go on and on. I'm not sure what happens at the limit, though.
Therefore, this raises a nice question: