Is it okay to do ANOVA on group that is compared with itself and the rest? I have 4 clusters that is obtained from clustering i.e. Group A, Group B, Group C, and Group D. Now, the comparison that other groups had done was Group A vs. Rest, Group B vs Rest, Group C vs Rest, Group D vs Rest. They're asking me to do ANOVA to find any differences between the groups. According to them, the rows in the groups were oversampled i.e. some rows in Group A could also be in Group B or Group C or in both, for example. Is it even okay to do this? Then is it also okay to perform post-hoc comparison? Could you please help me understand this?
EDIT
The shape of the Dataframe with ties is listed below. The data is joined on 'ID':
Merged data (Inner-Join)
=========================

Comparison              Shape           
1. Group A vs Group B:  (15, 107)
2. Group A vs Group C:  (0, 107)
3. Group A vs Group D:  (0, 107)
4. Group B vs Group C:  (15, 107)
5. Group B vs Group D:  (0, 107)
6. Group C vs Group D:  (16, 107)

Actual total rows: 211
Total rows after oversampling: 270

 A: Data for illustration: Here are some fake data (simulated in R) to use for illustration: four groups, each with 30
observations, common population standard deviation $\sigma = 3,$ and various
population means. (There are several ANOVA procedures in R. In some of them, it is essential to make gp a factor variable.)
set.seed(1234); n = 30; sg = 3
a = rnorm(n, 20, sg)
b = rnorm(n, 20, sg)
c = rnorm(n, 25, sg)
d = rnorm(n, 35, sg)
x = c(a,b,c,d);  gp = as.factor(rep(1:4, each=n))

In 'notched' boxlots, the notches in the sides of the boxes are calibrated
for making nonparametric comparisons between two groups at a time. If notches overlap between any two groups, then those two groups may not be significantly different.
boxplot(x ~ gp, col="skyblue2", pch=20, notch=T)


Tentatively, it seems that groups 1 and 2 may have the same population means, and that
groups 3 an 4 may have different means, both greater than in groups 1 and 2.
ANOVA Table and main F-test: The one-way ANOVA table is made as follows:
aov.out = aov(x ~ gp)
summary(aov.out)
             Df Sum Sq Mean Sq F value Pr(>F)    
gp            3   5554  1851.4   245.8 <2e-16 ***
Residuals   116    874     7.5                   
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

With such a small P-value, there is strong evidence that not all four group
population means are equal.
Multiple comparisons: There are many methods of adjusting P-values for pairwise comparisons of
group sample means in order to avoid false rejections. Below I show the
results from the Bonferroni method. You can look online for several other
methods.
pairwise.t.test(x, gp, p.adj="bon")

        Pairwise comparisons using t tests with pooled SD 

data:  x and gp 

  1       2       3      
2 1       -       -      
3 5.4e-14 < 2e-16 -      
4 < 2e-16 < 2e-16 < 2e-16

P value adjustment method: bonferroni 
Group 1 does differ significantly from group 2 (P-value 1), but does
differ significantly from the other groups (P-values near 0).
Group 2 differs significantly from both of the other groups, which in
turn differ significantly from each other.
Potential design errors in your data: From a vague description of 'oversampling' that 'may have' taken place, it is
difficult to know what effect such design errors may have on multiple comparisons. 
If a few Group A values were repeated in other groups, that
would tend to make the groups look more alike than they really might be.
So you may be safe in believing the significant differences you find in
your data. However, if such repetition is extremely widespread it may make
the variance estimate too small which might lead to unwarranted deflation
of P-values.  
If values are expressed to several decimal places, you might
see what happens if you look for ties between the groups. Presumably, there will be few naturally occurring ties, and presumably repeated values would show as ties.
In my fake data, there is no such 'oversampling'. If I round my 120 observations to three place accuracy, there are no ties. (If I round to two
places, there are four ties among the 120 rounded values.)
x1 = round(x, 3)
length(unique(x1))
[1] 120

If you have few tied data (just speculating, say fewer than 5%), you might consider removing all
tied observations from your data and seeing if the that makes any
difference in what is declared significant.
