Wilcoxon (WMWU) test sensitivity Statistics courses often give the Wilcoxon-Mann-Whitney U test ("Wilcoxon" from now on) as an alternative to the two-sample t-test.
However, the test is not quite the test of medians that would be so convenient to call a robust alternative to the t-test, and I have read that it can be sensitive to general distribution differences.
I have not been able to get sensitivity to differences in shape or spread. Could someone please give an example of distributions with equal location (I mostly mean median but will leave this vague) where Wilcoxon should have unusually high power to reject?
 A: I suppose we need distributions of remarkably different shape.
For example, $\mathsf{Beta}(15,15)$ and $\mathsf{Beta}(.2,.2)$ both have means
and medians of $1/2.$ 
par(mfrow=c(1,2))
  curve(dbeta(x, 15, 15), 0, 1, lwd=2, ylab="Density", 
        main="BETA(15,15)")
    abline(h=0, col="green2")
  curve(dbeta(x, .2, .2), 0, 1, ylim=c(0,4), lwd=2, ylab="Density", 
        main="BETA(.2,.2)")
   abline(h=0, col="green2")
par(mfrow=c(1,1))


But the 2-sample Wilcoxon test with 'significance level' 5%
rejects for about 10% of samples of size $n =50.$
set.seed(2020)
pv = replicate(10^5, wilcox.test( rbeta(50,15,15), 
                                  rbeta(50,.2,.2) )$p.val)
mean(pv <= .05)
[1] 0.0976

If this were a true test of equal medians, then a histogram
of 100,000 P-values should be approximately uniform.
hist(pv, prob=T, col="skyblue2", main="Non-Uniform P-Values")

 
This was my first experiment. Maybe you can find a pair of distributions
with a stronger effect.
Addendum: Second experiment with (asymmetrical) exponential distributions.
If $X \sim \mathsf{Exp}(1),$ then the median of $X$ is 
$\eta = -\log(1/2), \approx 0.693.$ So 
$Y \sim \mathsf{Exp}(\mathrm{rate}=\eta)$ has median $1.$ and $Z = Y-1$ has median $0,$ as does $-Z.$ 
Now, let's see what happens if we use the Wilcoxon RS test
to distinguish between samples of size $n = 100$ from the distributions
of $Z$ and, independently, $-Z.$ A Wilcoxon RS test at the 5% level
rejects with probability nearly $2/3.$ 
set.seed(610);  h = -log(.5)
pv = replicate(10^5, wilcox.test( rexp(100,h) - 1, 
                                  1 - rexp(100,h) )$p.val)
mean(pv <= .05)
[1] 0.66367

hist(pv, prob=T, col="skyblue2", main="Non-Uniform P-Values")


A: One systematic way to do this is to take a location-scale family with an asymmetric distribution, so that the median depends on both location and scale. Center the two distributions at their medians; their means and other summary statistics will differ. 
For example, take shifted exponentials
r1<-replicate(10000,{
    x<-rexp(100,1)-log(2)
    y<-rexp(100,2)-log(sqrt(2))
    median(x)-median(y)
})

> summary(r1)
      Min.    1st Qu.     Median       Mean    3rd Qu.       Max. 
-0.3568476 -0.0738395 -0.0009061  0.0024136  0.0747878  0.4667098 


and
r<-replicate(10000,{
    x<-rexp(100,1)-log(2)
    y<-rexp(100,2)-log(sqrt(2))
    wilcox.test(x,y)$p.value
})
hist(r)


Another systematic way is to transform and shift a distribution, eg
> r<-replicate(10000,{
+     x<-rnorm(100)
+     y<-exp(rnorm(100))-exp(0)
+     wilcox.test(x,y)$p.value
+ })
 hist(r,col="red")


