The implication $ACC \implies UCC$ is irreversible in $ZF$. This follows again from the transfer theorem of Pincus:
$UCC$ is an injectively boundable statement, see note 103 in "Consequences of the axiom of choice" by Howard & Rubin. Right after the theorem in pp. 285 and its corollary, there are examples of statements of this kind, one of which is form 31 (which is precisely $UCC$). The fact that this is the case follows in turn from the application of lemma 3.5 in Howard, P.-Solski, J.: "The strength of the $\Delta$-system lemma", Notre Dame J. Formal Logic vol 34, pp. 100-106 - 1993
It is known that $¬ACC$ is boundable, and hence injectively boundable.
Then we can apply the transfer theorem of Pincus to the conjunction $UCC \wedge ¬ACC$ and we are done.
SECOND PROOF: Browsing through the "Consequences..." book I've just found another less direct proof of the same fact. I'm adding it here to avoid the interested person from looking it up for himself (especially since the AC website is not working these days). It involves form 9, known as $W_{\aleph_0}$: a set is finite if and only if it is Dedekind finite. Now, $ACC \implies W_{\aleph_0}$ is provable in $ZF$ (in fact this was already proved by Dedekind), while $UCC$ does not imply $W_{\aleph_0}$. The latter follows from the fact that in the basic Fraenkel model, $\mathcal{N}1$, $UCC$ is valid while $W_{\aleph_0}$ is not, and such a result is transferable by considerations of Pincus that can be found at Pincus, D.: "Zermelo-Fraenkel consistency results by Fraenkel Mostowski methods", J. of Symbolic Logic vol 37, pp. 721-743 - 1972 (doi:10.2307/2272420)
David Pincus in "A note on the cardinal factorial" (Fundamenta Mathematicae vol.98(1), pages 21-24(1978)) proves that $A^A=2^A$ does not imply the axiom of choice, therefore it does not characterise the sets for which $A=A\times A$. The counterexample is the model from his paper "Cardinal representatives", Israel Journal of Mathematics, vol.18, pages 321-344 (1974).
As Pincus writes on the last lines of [Pincus78]:
c. Our arguments [ in the proof of "4.The Main Theorem; If $x$ is infinite then $2^x=x!$" ] have made little use of the particular definition of $x!$. Indeed let $\mathcal{F}$ be any set valued operation which satisfies:
(1) The predicate $y\in\mathcal{F}$ is absolute from $M$ to $V$.
(2) $\mathsf{ZF}$ proves $|y|\leq x \Rightarrow |\mathcal{F}(y)|\leq |\mathcal{F}(y)|$ and $|2x|=|x|\Rightarrow 2^x\leq|\mathcal{F}(x)|$ for infinite $x$.
(3) $\mathsf{ZFC}$ proves $2^x=|\mathcal{F}(x)|$ for infinite $x$.
The statement "For every infinite $x$, $2^x=|\mathcal{F}(x)|.$", holds
in $M$ (and therefore is not an equivalent to the axiom of choice).
Examples of $\mathcal{F}$, apart from $x!$, are $x^x$ and $x^x-x!$.
Therefore, in this Pincus model, $\mathsf{ZF}$ + $\lnot\mathsf{AC}$ + "for all infinite x, $2^x=x!=x^x=x^x-x!=|\mathcal{F}(x)|$" holds for any set valued operator $\mathcal{F}$ as above.
I should say that, after failing to come up with an answer myself, I found out about this by searching into my good old friend the "Consequences of the axiom of choice" by Howard and Rubin, Form 200, and Note 64. I try to reference this book whenever I can :)
Best Answer
It is shown in
Tachtsis, E.
On the existence of permutations of infinite sets without fixed points in set theory without choice.
Acta Math. Hungar. 157 (2019), no. 2, 281-300.
that ZF+(every infinite set supports a permutation with no fixed points) does not imply AC. It is easy to see in ZF that a finite set supports a cyclic permutation, so that should be sufficient for your question.