(?!\d)

'\.(\d)(\D|$)', '\.0\1'

with test as (select '1.1' code
from dual union all select  '1.2'
from dual union all select '1.11'
from dual)
SELECT code, REGEXP_REPLACE( code, '\.(\d)(\D|$)', '.0\1')
FROM test;

CODE	REGEXP_REPLACE(CODE,'\.(\D)(\D|$)','.0\1')
1.1	1.01
1.2	1.02
1.11	1.11

The following example examines country_name. Oracle puts a space after each non-null character in the string.

SELECT
  REGEXP_REPLACE(country_name, '(.)', '\1 ') "REGEXP_REPLACE"
  FROM countries;

REGEXP_REPLACE
--------------------------------------------------------------------------------
A r g e n t i n a
A u s t r a l i a
B e l g i u m
B r a z i l
C a n a d a
. . .

The following example examines the string, looking for two or more spaces. Oracle replaces each occurrence of two or more spaces with a single space.

SELECT
  REGEXP_REPLACE('500   Oracle     Parkway,    Redwood  Shores, CA',
                 '( ){2,}', ' ') "REGEXP_REPLACE"
  FROM DUAL;

REGEXP_REPLACE
--------------------------------------
500 Oracle Parkway, Redwood Shores, CA

> with t as (select '1.2.3.4.5.6.7.8.9.0' str from dual
)
select str 
     , regexp_replace(str||'.','\.(\d)\.','.0\1.') m_str
     , rtrim(
       regexp_replace(
         regexp_replace(str||'.','\.(\d)\.','.0\1.')
         ,'\.(\d)\.','.0\1.'),'.') m2_str
from t
union all
select str 
     , regexp_replace(str,'\.(\d)(\D|$)','.0\1\2') m_str
     , regexp_replace(
         regexp_replace(str,'\.(\d)(\D|$)','.0\1\2')
         ,'\.(\d)(\D|$)','.0\1\2') m2_str
from t         
STR                 M_STR                     M2_STR                       
------------------- ------------------------- ------------------------------
1.2.3.4.5.6.7.8.9.0 1.02.3.04.5.06.7.08.9.00. 1.02.03.04.05.06.07.08.09.00   
1.2.3.4.5.6.7.8.9.0 1.02.3.04.5.06.7.08.9.00  1.02.03.04.05.06.07.08.09.00   

> with t as (select '1.2.3.4.5.6.7.8.9.0' str from dual
)
select str 
     , ltrim(regexp_replace(str,'\D?(\d)(\D|$)','0\1\2'),'0') m_str
from t         
STR                 M_STR                        
------------------- ------------------------------
1.2.3.4.5.6.7.8.9.0 1.02.03.04.05.06.07.08.09.00   

> with t as (select '1.22.3.4.5.63.7.8.9.0' str from dual
)
select str 
     , replace(regexp_replace(replace(str,'.','..'),'\.(\d)(\.|$)','.0\1'),'..','.') m_str
from t         
STR                   M_STR                        
--------------------- ------------------------------
1.22.3.4.5.63.7.8.9.0 1.22.03.04.05.63.07.08.09.00