It is informative to see exactly what the Mann-Whitney test does. For two samples $X = \{x_1, \dots, x_m \}$ and $Y=\{y_1, \dots, y_n\}$, under the assumptions that
- Observations in $X$ are iid
- Observations in $Y$ are iid
- The samples $X$ and $Y$ are mutually independent.
- The respective populations from which $X$ and $Y$ were sampled are continuous.
then, the U statistic is defined as:
$$ U = \sum_{i=1}^m \sum_{j=1}^n bool(x_i < y_j )$$
It should be reasonably intuitive to see that if X and Y represent the same distributions (i.e. the null hypothesis), then the expected value of $U$ would $mn/2$, since you could expect values below a certain rank to occur as often for $X$ as for $Y$. So you can think of the Mann Whitney test as checking to what extent the statistic $U$ deviates from this expected value.
If this intuition isn't clear, then think of the first rank (i.e. the leftmost rarest value in each sample). If $X$ and $Y$ were drawn from the same distribution, you would have no reason to expect that the rarest value in $X$ would be less than $Y$ more than 50% of the time, otherwise this would make you think that actually $X$ has a heavier tail than $Y$. You can extend this logic for the 2nd rarest value, 3rd, and so forth.
Similarly, if you drew the same number of observations, say $K$, you could almost think of the ranks as $K$ "common bins" with fuzzy boundaries. If $X$ and $Y$ came from the same population, you might expect each rank to occupy roughly the same space, and there's no reason to think that the $x_k $observation in that bin would be to the right of $y_k$ more than 50% of the time.
However, if $x_k$ at a particular "bin" $k$ was to the right of $y_k$ more often than not, then this denotes that there is a systematic "shift". This is what makes Mann-Whitney a good test for detecting 'shift' in distributions that are assumed to be relatively similar except for a possible shift due to a treatment effect.
Now consider the $X \sim \mathcal N(0,1)$ vs $Y \sim \mathcal N(0,2)$ scenario. Assume $K=1000$ samples in each case. You would expect that for the most part, given the same rank, negative values in Y, would tend to be to the left of X more or less all the time. Whereas, positive values in Y, would tend to be to the right of X more or less all the time. Therefore in this particular scenario, even though the distributions are completely different, it happens that half the time X is less likely to be larger than Y, and half the time it is more likely. Therefore you'd expect the U statistic to be very close to the expected value $K^2/2$, and therefore unlikely to be significant.
In other words, it may be a reasonable test to compare two samples in a general "goodness of fit" sense in some specific circumstances, but it is important to be familiar with the situations where it would not. The example above is one such case.
Best Answer
You could use the fact that the distribution of the sample variance is a chi square distribution centered at the true variance. Under your null hypothesis, your test statistic would be the difference of two chi squared random variates centered at the same unknown true variance. I do not know whether the difference of two chi-squared random variates is an identifiable distribution but the above may help you to some extent.