Wilcoxon Test - non normality, non equal variances, sample size not the same I know there are already a lot of posts out there, but I couldn't find this exact combination in any of them. 
Comparing two samples (Prices associated with men and with women), but I have neither the same sample size ($n = 790$ vs $n=795$) nor equal variance or normality. 
My hypothesis is whether the prices for Women are greater than prices for Men. The Wilcoxon test is significant on a 4 % level. 
Can I actually say anything helpful since so many assumptions are violated? Would another test be better?
EDIT Some additional infos: 
Prices Women: Median 28.00, Mean 28.47,  Std. Dev. 17.17,  Skewness 0.91, Kurtosis 5.41
Prices Men:   Median 26.00, Mean, 29.08, Std. Dev. 22.43,  Skewness 2.39, Kurtosis 12.74

 A: tl;dr if you want to interpret the rejection of the null hypothesis as evidence that prices for women are greater than those for men, then you do need the assumption of equal variance (in fact, equal distributions) between the two populations. If you are satisfied with showing that the distribution of prices for women differs in some way from that of men, then you don't need the extra assumption.
You don't need to worry about unequal sample size (this will affect the power of the test, but not its validity) or Normality.
For what it's worth, testing whether one group's values are larger on average than another group's when their variances also differ is a surprisingly deep question, even for Normally distributed data (where it's known as the Behrens-Fisher problem).
Referring to the Wikipedia page: the "very general formulation" says:


  
*Under the null hypothesis H0, the distributions of both populations are equal.[3]
  
*The alternative hypothesis H1 is that the distributions are not equal.
  

The next paragraph says: 

Under more strict assumptions than the general formulation above, e.g., if the responses are assumed to be continuous and the alternative is restricted to a shift in location, i.e., $F_1(x) = F_2(x + δ)$, we can interpret a significant Mann–Whitney U test as showing a difference in medians ...

(emphasis added)

Note to technical readers: I think this is a reasonable summary, but if anyone wants to be more rigorous, feel free to comment or edit or post an alternative answer ...
A: Ben Bolker's answer is great. I just wanted to add an answer to the "Would another test be better?" part.
If you want to be able to conclude that prices for women are greater than those for men without the assumptions of equal distributions under the null, Brunner-Munzel's test can be recommended. For a full technical description of this test, see Chapter 3 in https://link.springer.com/content/pdf/10.1007/978-3-030-02914-2.pdf. For a non-technical introduction and comparison to the Wilcoxon test, see https://journals.sagepub.com/doi/full/10.1177/2515245921999602 (Disclaimer: I am the author of this one).
A disclaimer: this test is not optimal in any strict sense. However, it seems to be the most reasonable test for the nonparametric Behrens-Fisher problem available at the moment, just as Welch's t test is not optimal in any sense but most reasonable if you want to test equal means. Indeed, Brunner-Munzel's test is almost Welch's t test on ranked data, while the Wilcoxon test is close to Student's t test on ranked data.
