Generate Dennis Numbers This challenge is a tribute to PPCG user Dennis for winning the robbers' part of The Programming Language Quiz.
Looking at Dennis' PPCG profile page we can see some pretty impressive stuff:

He currently has over sixty-eight thousand reputation, making him second in rep overall, surpassing third place by almost thirty thousand. He recently won our election for a new moderator and got a shiny new diamond next to his name. But I personally think the most interesting part about Dennis is his PPCG user ID number: 12012.
At first glance 12012 almost looks like a palindrome, a number that reads the same when reversed, but it's a little off. It can become the palindrome 21012 if we swap the positions of the first 1 and 2, and it can become the palindrome 12021 if we swap the last 1 and 2. Also, following the convention that leading zeroes in a number are not written, swapping the first 1 and the 0 results in 02112 or rather 2112 which is another palindrome.
Let's define a Dennis number as a positive integer that is not palindromic itself but can be made into a palindrome by swapping the positions of at least one pair of any two digits. The order of a Dennis number is the number of distinct pairs of digits that can be swapped to make a (not necessarily distinct) palindrome.
So the order of 12012 is 3 since 3 distinct pairs of its digits (12012, 12012, 12012) can be swapped around to produce palindromes. 12012 happens to be the smallest order 3 Dennis number.
10 is the smallest Dennis number and has order 1 because switching around the 1 and 0 gives 01 a.k.a. 1 which is a palindrome.
The imaginary leading zeroes of a number don't count as switchable digits. For example, changing 8908 to 08908 and swapping the first two digits to get the palindrome 80908 is invalid. 8908 is not a Dennis number.
Non-Dennis numbers could be said to have order 0.
Challenge
Write a program or function that takes in a positive integer N and prints or returns the Nth smallest Dennis number along with its order in some reasonable format such as 12012 3 or (12012, 3).
For example, 12012 is the 774th Dennis number so if 774 is the input to your program, the output should be something like 12012 3. (Curiously, 774 is another Dennis number.)
The shortest code in bytes wins.
Here is are the first 20 Dennis numbers and their orders for reference:
N       Dennis  Order
1       10      1
2       20      1
3       30      1
4       40      1
5       50      1
6       60      1
7       70      1
8       80      1
9       90      1
10      100     1
11      110     2
12      112     1
13      113     1
14      114     1
15      115     1
16      116     1
17      117     1
18      118     1
19      119     1
20      122     1

Here is the same list up to N = 1000.
CJam, 45 bytes

0{{)_s:C,2m*{~Ce\is_W%=},,2/:O!CCW%=|}g}ri*SO


Try it online!

How it works

0          e# Push 0 (candidate).
{          e# Loop:
  {        e#   Loop:
    )_     e#     Increment the candidate and push a copy.
    s:C    e#     Cast to string and save in C.
    ,      e#     Get the length of C, i.e., the number of digits.
    2m*    e#     Push all pairs [i j] where 0 ≤ i,j &lt; length(C).
    {      e#     Filter:
      ~    e#       Unwrap, pushing i and j on the stack.
      Ce\  e#       Swap the elements of C at those indices.
      is   e#       Cast to int, then to string, removing leading zeroes.
      _W%= e#       Copy, reverse and compare.
    },     e#     Keep the pairs for which = returned 1, i.e., palindromes.
    ,2/    e#     Count them and divide the count by 2 ([i j] ~ [j i]).
    :O     e#     Save the result (the order) in O.
    !      e#     Negate logically, so 0 -&gt; 1.
    CCW%=  e#     Compare C with C reversed.
    |      e#     Compute the bitwise NOT of both Booleans.
           e#     This gives 0 iff O is 0 or C is a palindrome.
  }g       e#   Repeat the loop while the result is non-zero.
}ri*       e# Repeat the loop n times, where n is an integer read from STDIN.
           e# This leaves the last candidate (the n-th Dennis number) on the stack.
SO         e# Push a space and the order.