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.
In the following answer, by Foreman-woodin model, I mean the model constructed by them in the paper "The generalized continuum hypothesis can fail everywhere.
Ann. of Math. (2) 133 (1991), no. 1, 1–35. "
Questions 1 and 3 have positive answer: In Foreman-Woodin model for the total failure of GCH the following hold:
1) For all infinite cardinal $\kappa, 2^{\kappa}$ is weakly inaccessible, and hence a fixed point of the $\aleph-$function,
2) If $\kappa \leq \lambda< 2^{\kappa},$ then $ 2^{\lambda}= 2^{\kappa}.$
In this model for all infinite cardinals $\kappa, 2^{\kappa}=\aleph_{ 2^{\kappa}}$ in particular for all fixed points $\kappa$ of the $\aleph-$function, $2^{\aleph_\kappa}=\aleph_{ 2^{\kappa}}$. Also note that in this model for all infinite cardinals $\kappa,$
if we let $\lambda=2^{\aleph_\kappa},$ then $\lambda \geq 2^\kappa,$ and $2^{\aleph_\kappa}=\aleph_\lambda.$ So both of questions 1 and 3 have a positive answer.
For your question 2, $\delta$ can be arbitrary large: Start with GCH+there exists a supercompact cardinal $\kappa$+ there are infinitely many inaccessibles above it. Now let $\delta$ be any ordinal $<\kappa.$ Force with Foreman-Woodin construction above $\delta$ (in the sense that let the first point of the Radin club added during their forcing construction be above $\delta$). In their final model (which is $V_\kappa$ of some extension of the ground model) for all infinite cardinals $\lambda <\kappa, 2^\lambda \geq \lambda^{+\delta}$. So if $F$ is defined in the ground model by $F(\kappa)=\kappa^{+\delta},$ then the acceleration rank of $F$ is $\delta$ (using GCH), and in the finial model for all infinite cardinals $\kappa, 2^\kappa \geq F(\kappa).$
Remark. I may note that we can not define the function $F$ in the ground model, such that it is the realization of power function in the extension, but we can find some inner model of the final extension in which $GCH$ holds and such a function $F$ is definable.
Best Answer
The answer is positive, yes.
Note that $2^\kappa\leq 2^{\kappa^+}$, and therefore $\kappa^+\leq\kappa^++2^\kappa\leq 2^{\kappa^+}$. So either $2^\kappa=2^{\kappa^+}$ or $2^\kappa=\kappa^+$.
In the first case $\kappa^+$, $\kappa<\kappa^+<2^\kappa$ is impossible. So the latter case holds.
Therefore the power set of an ordinal can be well-ordered, and the Axiom of Choice follows in $\sf ZF$.