You need to apply an itertools-style-groupby and then keep the rows where Y is maximal.

>>> df 
   idx  X   Y
0    0  A   3
1    1  B   2
2    2  A   7
3    3  A  10
4    4  B   1
5    5  C   4
6    6  A   3
7    7  A   5
>>> y_max = df.groupby(df['X'].ne(df['X'].shift()).cumsum())['Y'].transform('max')
>>> df[df['Y'] == y_max] 
   idx  X   Y
0    0  A   3
1    1  B   2
3    3  A  10
4    4  B   1
5    5  C   4
7    7  A   5

edit:

Initial solution had a bug and only produced the correct idx column by accident.

edit 2:

If you only want to keep one row per group, you can use

>>> y_idxmax = df.groupby(df['X'].ne(df['X'].shift()).cumsum())['Y'].idxmax()
>>> df.loc[y_idxmax] 
   idx  X   Y
0    0  A   3
1    1  B   2
3    3  A  10
4    4  B   1
5    5  C   4
7    7  A   5

Credit goes to Ch3steR for this one.

Or I'd prefer just simply only specify the groups in the groupby parameters:

df.groupby(df['X'].ne(df['X'].shift()).cumsum(), as_index=False).max()

Or:

df.groupby(df['X'].ne(df['X'].shift()).cumsum()).max().reset_index(drop=True)

Both output:

   idx  X   Y
0    0  A   3
1    1  B   2
2    3  A  10
3    4  B   1
4    5  C   4
5    7  A   5

Create a column which heaps consecutives into one group

   df['temp']=(~(df['X']==df['X'].shift())|(df['X'].shift(-1)==df['X'])).cumsum()

groupby the group of consecutives and filter out wherevalues of Y are equal to maximum in each group. Drop the temp column

df[df.groupby('temp')['Y'].transform(lambda x:(x==x.max()))].drop(columns=['temp'])

A neater way is not to create column but save the groups of consecutives into a variable and groupby the variable as follows

s=(~(df['X']==df['X'].shift())|(df['X'].shift(-1)==df['X'])).cumsum()
print(df[df.groupby(s)['Y'].transform(lambda x:(x==x.max()))])

    idx  X   Y
0    0  A   3
1    1  B   2
3    3  A  10
4    4  B   1
5    5  C   4
7    7  A   5