If I understand you correctly, you are describing the following:

Setup

d = {'id1': ['x', 'x'], 'id2': ['z','w'], 'metric': [100,10] }
df = pd.DataFrame(data=d)
df

Solution

# Manually choose the value by which to scale the column 'metric'
scaler = df.loc[(df['id1'] == 'x') & (df['id2'] == 'z'), 'metric'].values

# Divide all 'metric' values by the above scaler value
df['result'] = df['metric'] / scaler

df
  id1 id2  metric  result
0   x   z     100     1.0
1   x   w      10     0.1

Create all pairs combinations of columns names, loop and divide to new columns:

from  itertools import combinations

for a, b in combinations(df.columns, 2):
    df[f'{a}/{b}'] = df[a].div(df[b])

Or use list comprehension, join together by concat and add original columns by join:

df = df.join(pd.concat([df[a].div(df[b]).rename(f'{a}/{b}') 
                         for a, b in combinations(df.columns, 2)], 1))

print (df)
   x1  x2  x3  x4     x1/x2     x1/x3     x1/x4     x2/x3     x2/x4     x3/x4
0   4   7   1   5  0.571429  4.000000  0.800000  7.000000  1.400000  0.200000
1   5   8   3   3  0.625000  1.666667  1.666667  2.666667  2.666667  1.000000
2   4   9   5   6  0.444444  0.800000  0.666667  1.800000  1.500000  0.833333
3   5   4   7   9  1.250000  0.714286  0.555556  0.571429  0.444444  0.777778
4   5   2   1   2  2.500000  5.000000  2.500000  2.000000  1.000000  0.500000
5   4   3   0   4  1.333333       inf  1.000000       inf  0.750000  0.000000

We can take advantage of the way that Boolean values are handled mathematically (True being 1 and False being 0) and use 3 aggregation functions sum, count and mean per group (groupby aggregate). We can also take advantage of Named Aggregation to both create and rename the columns in one step:

df = (
    df.groupby('playerId', as_index=False)
        .agg(wins=('winner', 'sum'),
             totalCount=('winner', 'count'),
             winPct=('winner', 'mean'))
)
# Scale up winPct
df['winPct'] *= 100

df:

   playerId  wins  totalCount  winPct
0      1848     1           2    50.0
1      1988     0           2     0.0
2      3543     1           1   100.0

DataFrame and imports:

import pandas as pd

df = pd.DataFrame({
    'playerId': [1848, 1988, 3543, 1848, 1988],
    'winner': [True, False, True, False, False]
})

This should do:

df['price_per_sqft'] = df['SalePrice']/df['LotArea']

or you can use pd.assign

df.assign(price_per_sqft = df.SalePrice/df.LotArea)

How about:

user_count=df3.groupby('user_state')['user_count'].mean()
#(or however you think a value for each state should be calculated)

engaged_unique=df3.groupby('user_state')['engaged_count'].nunique()

engaged_pct=engaged_unique/user_count

(you could also do this in one line in a bunch of different ways)

Your original solution was almost fine except that you were dividing a value by the entire user count series. So you were getting a Series instead of a value. You could try this slight variation:

def f(x):
    engaged_percent = x['engaged_count'].nunique()/x['user_count'].mean()
    return engaged_percent

by = df3.groupby(['user_state']).apply(f)
by

After performing your groupby, use pct_change:

# Sort the DataFrame, if necessary.
df = df.sort_values(['name', 'order'])

# Use groupby and pcnt_change on the 'quantity' column.
df['quantity'] = df.groupby('name')['quantity'].pct_change()

The resulting output:

  name  order  quantity
0    A      1       NaN
1    A      2  0.500000
2    A      3 -0.666667
3    B      1       NaN
4    B      2  2.000000

Idea is create MultiIndex and then add one year for datetimes in rename, divide columns and add suffix by DataFrame.add_suffix, last add to original by DataFrame.join:

df1 = df.set_index(['Date','Route'])

df2 = df1.rename(lambda x: x + pd.offsets.DateOffset(years=1), level=0)

df3 = df1[['Col1','Col2']].div(df2[['Col1','Col2']]).add_suffix('_ratio')

df = df.join(df3, on=['Date','Route'])
print (df)
         Date Route  Col1  Col2  Col1_ratio  Col2_ratio
0  2020-01-01     A    99    26         NaN         NaN
1  2020-01-02     A    37    96         NaN         NaN
2  2020-01-03     A    20    83         NaN         NaN
3  2021-01-01     A    50    79    0.505051    3.038462
4  2021-01-02     A    16    50    0.432432    0.520833
5  2021-01-03     A    52    33    2.600000    0.397590
6  2020-01-01     B    48    44         NaN         NaN
7  2020-01-02     B    96    30         NaN         NaN
8  2020-01-03     B    17    42         NaN         NaN
9  2021-01-01     B    34    74    0.708333    1.681818
10 2021-01-02     B    49    37    0.510417    1.233333
11 2021-01-03     B    70    12    4.117647    0.285714

You can use custom function f with dt.days for convert Timedelta to daysand divide by div:

def f(x):
    d = x.date.diff().dt.days
    q = x.quantity.diff()
    return (q.div(d)).mean()

df1 = df.sort_values('date').groupby('name').apply(f).reset_index(name='ratio')
print (df1)
  name  ratio
0  'A'   -5.0
1  'B'    NaN