For finitedimensional (nonweak) Hopf C*algebras it is known that the antipode is always involutive, as claimed e.g. in https://arxiv.org/pdf/1007.5283.pdf. I couldn't find the same statement for weak Hopf C*algebras, however. Are there counterexamples?
2 Answers
According to this paper of NikshychVainerman, a finitedimensional weak Kac algebra is precisely a finitedimensional weak Hopf ${\rm C}^{\star}$algebra with an involutive antipode ($S^2 = id$), and they give (p306) an explicit example of finitedimensional weak Hopf ${\rm C}^{\star}$algebra which is not a weak Kac algebra.
More generally, any finite index depth $2$ unital inclusion of type ${\rm II}_1$ factors is characterized by a finitedimensional weak Hopf ${\rm C}^{\star}$algebra and in the weak Kac case, the index is an integer. So every noninteger finite index depth $2$ subfactor provides an example of finitedimensional weak Hopf ${\rm C}^{\star}$algebra with a noninvolutive antipode. For any depth $n$ subfactor $(N \subset M)$ of finite index $M:N$, if $$N \subset M \subset M_1 \subset M_2 \subset M_3 \subset \cdots $$ is the Jones tower, then $(N \subset M_{n2})$ is a depth $2$ subfactor of index $M:N^{n1}$. The depth $n$ TemperleyLiebJones subfactor has index $4cos^2(\frac{\pi}{n+1})$, and $[4cos^2(\frac{\pi}{n+1})]^{n1}$ is not an integer for $n \neq 2,4$, so the corresponding weak Hopf ${\rm C}^{\star}$algebra has a noninvolutive antipode (it should also be the case for $n=4$, but not for $n=2$). The example of NikshychVainerman above is $n=3$.
This paper of Das proves by planar algebra techniques that for any finite index depth $2$ subfactor $(N \subset M)$, the relative commutants $N' \cap M_1$ and $M' \cap M_2$ has a structure of finitedimensional weak Hopf ${\rm C}^{\star}$algebra. What does the diagram on p84 say about the antipode $S$?

$\begingroup$ The diagram p84 says that the antipode $S$ is a contragredient up to conjugation (i.e. $S(a) = w^{1} \overline{a} w$), and is involutive up to conjugation (because $S^2(a) = w^{2} a w^2$). $\endgroup$ Jan 16, 2018 at 6:27

$\begingroup$ Editing a dozen old questions in the space of a few minutes drives newer questions off the front page. Please don't do that. $\endgroup$ Feb 12, 2021 at 11:43

$\begingroup$ @GerryMyerson: Sorry about that. By the way, what are you doing at the front page? $\endgroup$ Feb 12, 2021 at 12:30

$\begingroup$ I don't understand your question. When one goes to mathoverflow.net one lands on the front page. $\endgroup$ Feb 12, 2021 at 21:55

$\begingroup$ @GerryMyerson: sure, but do you go through all the questions on the front page (everyday)? If so, why? My personal use of MathOverflow is through "filter subscriptions" and newsletter emails. $\endgroup$ Feb 12, 2021 at 22:17
This is an answer to the question in the OPs answer.
I just took a brief look into
Rehren  Weak C Hopf symmetry* in Quantum Groups Symposium at "Group21", eds. H.D. Doebner et al., Goslar 1996 Proceedings, Heron Press, Sofia (1997), pp. 6269. https://arxiv.org/abs/qalg/9611007
which does the type III case.
The point is that $S^{1}(q)=S^\ast(q^\ast)$, which is what the diagram on p84 in the mentioned article says.