WITH s AS (
  SELECT
    t.id,
    string_agg(t.street, ' ' ORDER BY t.street) AS street
  FROM (
    SELECT 
      row_number() OVER () AS id,
      regexp_split_to_table(v.street, '\s+') AS street
    FROM 
      (
        VALUES 
          ('переулок Кедровый'), 
          ('Кедровый переулок')
      ) AS v(street)
    ) AS t
  GROUP BY
    t.id
)
SELECT
  s.id, s.street
FROM
  s
WHERE
  EXISTS (
    SELECT 1
    FROM s AS i
    WHERE
      i.id < s.id AND
      i.street = s.street
  )