Assumptions of Mann-Whitney test for at least ordinal data I am reading article by Divine et al. about using Mann-Whitney test for data that is at least ordinal (i.e. it may be discrete with many ties). It says the following (in section 2.3):

That is, it (the Mann-Whitney test) generally does not depend upon any particular distributional form (or parameters) in order to generate the test statistic and p-value. In fact, it is the whole distributions that are being compared, rather than any sample-specific summary statistic(s). However, the procedure does depend upon some assumptions about those distributions. For instance, one important assumption is that the variances of the two distributions should be the same (Pratt 1964).

And in section 5.1 this paper recommends to use the Brunner-Munzel test instead of the Mann-Whitney test if the variances are unequal (as well as scipy.stats.brunnermunzel manual):

Although the basic WMW test may be invalid with unequal
variances (especially with unequal sample sizes), the Brunner–Munzel variation should work if the minimum sample size is at least 30 and the variance discordance is not too extreme. For a sample size (or sizes) below 30 and/or when one or more large clumps of ties are present, an exact/permutation WMW test (available in SAS and R) should be considered.

The hypotheses in this article are formulated as follows (in the two-sided alternative case; $X_1 \sim F, X_2 \sim G$):

*

*$H_0: ~ P(X_1 \gt X_2) + \frac{1}{2} P(X_1 = X_2) = \frac{1}{2}.$

*$H_1: ~ P(X_1 \gt X_2) + \frac{1}{2} P(X_1 = X_2) \neq \frac{1}{2}.$
I am wondering what are the other assumptions of such Mann-Whitney test? (besides equality of variances and independence of samples; if we want to use this test for some at least ordinal data, i.e. not necessarily continuous)

In the famous article by Fay and Proschan (2010) there is a very similar formalization (perspective) of the Mann-Whitney test which is given for continuous data:

where $\Psi_C$ is the set of all continuous distributions, $H_3$ is the null and $K_3$ is the alternative, $\mathrm{P} = H_3 \sqcup K_3$ is the full set of allowed distributions.
The assumption of equal variances (which I mentioned earlier, see the beginning of this post) is one of the requirements which are introduced to guarantee that $\mathrm{P}$ will not contain distributions with both $\phi(F,G) = 1/2$ and $F \neq G$. And I want to know what are the other assumptions (besides equality of variances) we need to guarantee that.
Indeed, according to article by Karch (2021), "The assumptions for the different perspectives are all a special case of the Mann-Whitney test’s core assumption, exchangeability. In the Mann-Whitney test setting, exchangeability reduces to if the null hypothesis is true, the two population distributions must be identical." In other words, different perspectives have different null hypotheses but in each case the full set of the allowed distributions $\mathrm{P}$ shouldn't contain distributions $(F,G)$ for which it is possible to have $F \neq G$ under the null. That's why for each perspective we have different set of assumptions (i.e. restrictions on $\mathrm{P}$) to guarantee that.
Fay and Proschan require continuous distributions here (although they defined $\phi(F,G)$ both for discrete and continuous distributions). I guess that they require this because the consistency of Mann-Whitney test is strictly proved only for continuous distributions. However, the article by Divine et al. shows that the aforementioned  formalization of Mann-Whitney test (it is given at the beginning of my post as well as hyperlink to the article) is perfectly valid for discrete data (which possibly contain many ties).

There is some disagreement as to the  'proper' use
of the two-sample Wilcoxon (rank sum) test. Perhaps this is
because it is often used in ways that might surprise
its creators and because various software programs
have implemented a wide variety of versions to accommodate (moderate proportions of)
ties and other departures from canonical assumptions.
One way to be reasonably sure how the Wilcoxon RS test
works in a particular situation is to try it out and
see what actually happens.
The following brief simulations address the assumption
that the two populations must be of the same shape, differing
only by a shift; this assumption is often taken to mean
that the population variances must be equal.
By contrast, the implementation in R can be viewed as
a test whether one distribution stochastically dominates
the other--up to a point, regardless of shape or of variance.
I use the test to compare samples of size 50 from
distributions (a) $\mathsf{Norm}(\mu=100,\sigma=5),$
(b) $\mathsf{Norm}(\mu=100,\sigma=10),$ and (c)
$\mathsf{Norm}(\mu=105,\sigma=10).$
First, we use the Wilcoxon SR test to compare null (a) with alternative (b), a difference in shapes; second, to compare null (a) with alternative (c), a difference in shapes and locations.
set.seed(1123)
pv = replicate(10^4, wilcox.test(rnorm(50, 100, 5), 
                      rnorm(50,100,10))$p.val)
mean(pv <= .05)
[1] 0.0577         # (a vs b) true level about 6%, not exactly 5%

par(mfrow=c(1,3))
 hist(pv, prob=T, col="skyblue2", main="Same Centers")

pv = replicate(10^4, wilcox.test(rnorm(50, 100, 5), 
                      rnorm(50,105,10))$p.val)
mean(pv <= .05)
[1] 0.8483         # (a vs c) power about 85%

hist(pv, prob=T, br=20, col="skyblue2", main="Different Centers")

curve(pnorm(x,100,5),50,150, lwd=2, col="green3", lty="dashed")
 curve(pnorm(x,100,10), add=T, col="blue")
 curve(pnorm(x,105,10), add=T, col="maroon", lty="dotted")
par(mfrow=c(1,1))

The first panel of the figure shows the roughly uniform
distribution of of P-values of comparison (a) vs (b), and the second
shows the power (left-most histogarm bar) of comparison (a) vs (c).
The third panel shows that neither distribution (a) [broken green] nor (b) [solid blue] is stochastically dominant.
It also shows that (c) [dotted red] dominates (a), plotting mainly to the right of
and below (a).

Finally, we note that, because data are normal, the most appropriate
test to compare (a) and (b) would be a two-sample Welch t test, which does not assume equal variances; its significance
level is very near the nominal 5% level (no figure).
set.seed(1123)
pv = replicate( 10^4, t.test( rnorm(50, 100, 5), 
                       rnorm(50,100,10) )$p.val )
mean(pv <= .05)
[1] 0.0484      # aprx 5%

The point here is not to give an exhaustive catalog
of the properties of any one implementation of the Wilcoxon RS test. It is to illustrate how simple simulations can help to settle particular controversies.
Note: Original versions of the Wilcoxon rank sum test and the Mann-Whitney U test used different, but essentially equivalent, test statistics.
Addendum, per Comment.
If the task is to test whether $\mathsf{Beta}(1,3) \ne \mathsf{Beta}(3,1),$
based on ten observations from each distribution, then the two-sample Wilcoxon
test (2-sided) will do the job with power
very nearly 1:
set.seed(2022)
pv = replicate(10^5, wilcox.test(rbeta(10, 1,3), rbeta(10, 3,1))$p.val)
mean(pv <= 0.05)
[1] 0.99692

However, it seems that the meaning of rejection ('perspective') should not be that the median
of the former distribution is about $\eta_1=0.2063$ and $\eta_2=0.7937,$ and
even less that the median has "shifted" upward. The two distributions have very
different shapes.
It is clear from plots of empirical CDF
of two samples of size ten that $\mathsf{Beta}(3,1)$ (blue) dominates (tends to give larger values than) the former:
 set.seed(622)
 x1 = rbeta(10, 1, 3)
 x2 = rbeta(10, 3, 1)

 hdr="ECDF Plots: BETA(3,1) Dominates"
 plot(ecdf(x2), col="blue", xlim=0:1, main=hdr)
 plot(ecdf(x1), add=T, col="brown")


