user13597: Here is a knight rebus. Each letter represents a different type of piece.
Uppercase letters are one color and lowercase letters are the other.
Determine the position. If possible, also determine the previous moves.

n and N are kings and h is pawn. What is the rest?

user8201: This is a basic example of an orthodox retro form which has been taken to a new level over the last year by Ukrainian composer Andrey Frolkin and Canadian Jeff Coakley. I recently acquired a lovely landscape painting by Andrey's wife, Nina, which contains an embedded rebus by Andrey, who I bet authored the problem OP posted!
Although Maxwell's excellent proof relied at a couple of place on the edge of the board preventing retro-movement, this is not actually necessary. Shift all the pieces one square to the right (except for h on h2), and the problem remains sound (and in my opinion slightly superior :D )!
Proof: first count the pieces.

      K N I G H T
Upper 1 1 1 0 0 0
Lower 0 1 2 2 1 1

So N is king and by examining first and last ranks, H is pawn. Observe that whatever I represents, one instance is giving check to the enemy king. So any other checking unit implies a double check. So G is not a diagonal unit (bishop or queen), and K is not an orthogonal unit (rook or queen), since the resulting double checks are impossible. And I is not queen, or both kings would be in check!
So by elimination the queen is represented by T and we do have double check. So I can only be rook, and the position is achieved through promotion. Hence lower case is Black. G must be knight so by elimination K is bishop.  
[title "Frolkin? version Laska rebus: solution"]
[fen "2B5/8/2k5/1R6/2n5/3K4/1r5p/1q1r1n2 w - - 0 1"]

and Black's last move was c2xd1=R++

user4588: Yes indeed, n corresponds to king, as we need two kings of a different color to have a legal chess position.
Also, as you give, h = pawn, as otherwise we would have a pawn on the first or last rank, which would be illegal.
Consider i.
It cannot be a queen, as both kings would be in check.
It cannot be a knight, as the "i" on a2 would give check. There is no legal way where "knight a2 check" is the last move.
It cannot be a bishop, as the "I" on a5 would give check. There is no legal way where "bishop a5 check" is the last move.
Therefore, i = rook.
We see that the king on c3 is in check by the rook on c1. Therefore, g cannot be a queen or a bishop, because there is no legal way to obtain the resulting position.
So, g = knight.
The "K" on b8 should not give check to the king on b6.
Therefore, k = bishop and t = queen.
[However, the resulting position seems to be impossible to achieve too...
As far as I can see, the puzzle has no solution.]
Edit: the resulting position is indeed possible, see the answer of Glorfindel.

user7297: Assume the board is in it's usual position (the bottom row is the first row) then this is the solution:
[FEN "1B6/8/1k6/R7/1n6/2K5/r6p/q1r1n3 - - - 0 1"]

and Black's last move was b2xc1R++.
I could write an explanation here about I came to this result, but it would essentially be copying @Maxwell's answer. I didn't use it, though.

13597: <p>Here is a knight rebus. Each letter represents a different type of piece.
Uppercase letters are one color and lowercase letters are the other.</p>
<p>Determine the position. If possible, also determine the previous moves.</p>
<p><a href="https://i.stack.imgur.com/6JVio.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6JVio.png" alt="Rebus" /></a></p>
<p><strong>n</strong> and <strong>N</strong> are kings and <strong>h</strong> is pawn. What is the rest?</p>
 8201: <p>This is a basic example of an orthodox retro form which has been taken to a new level over the last year by Ukrainian composer Andrey Frolkin and Canadian Jeff Coakley. I recently acquired a lovely landscape painting by Andrey's wife, Nina, which contains an embedded rebus by Andrey, who I bet authored the problem OP posted!</p>

<p>Although Maxwell's excellent proof relied at a couple of place on the edge of the board preventing retro-movement, this is not actually necessary. Shift all the pieces one square to the right (except for h on h2), and the problem remains sound (and in my opinion slightly superior :D )!</p>

<p>Proof: first count the pieces.
<code>
      K N I G H T
Upper 1 1 1 0 0 0
Lower 0 1 2 2 1 1
</code></p>

<p>So N is king and by examining first and last ranks, H is pawn. Observe that whatever I represents, one instance is giving check to the enemy king. So any other checking unit implies a double check. So G is not a diagonal unit (bishop or queen), and K is not an orthogonal unit (rook or queen), since the resulting double checks are impossible. And I is not queen, or both kings would be in check!</p>

<p>So by elimination the queen is represented by T and we do have double check. So I can only be rook, and the position is achieved through promotion. Hence lower case is Black. G must be knight so by elimination K is bishop.  </p>

<pre><code>[title "Frolkin? version Laska rebus: solution"]
[fen "2B5/8/2k5/1R6/2n5/3K4/1r5p/1q1r1n2 w - - 0 1"]
</code></pre>

<p>and Black's last move was c2xd1=R++</p>
 4588: <p>Yes indeed, n corresponds to king, as we need two kings of a different color to have a legal chess position.</p>

<p>Also, as you give, h = pawn, as otherwise we would have a pawn on the first or last rank, which would be illegal.</p>

<p>Consider i.
It cannot be a queen, as both kings would be in check.
It cannot be a knight, as the "i" on a2 would give check. There is no legal way where "knight a2 check" is the last move.
It cannot be a bishop, as the "I" on a5 would give check. There is no legal way where "bishop a5 check" is the last move.
Therefore, i = rook.</p>

<p>We see that the king on c3 is in check by the rook on c1. Therefore, g cannot be a queen or a bishop, because there is no legal way to obtain the resulting position.
So, g = knight.</p>

<p>The "K" on b8 should not give check to the king on b6.
Therefore, k = bishop and t = queen.</p>

<p>[However, the resulting position seems to be impossible to achieve too...
As far as I can see, the puzzle has no solution.]</p>

<p><strong>Edit: the resulting position is indeed possible</strong>, see the answer of Glorfindel.</p>
 7297: <p>Assume the board is in it's usual position (the bottom row is the first row) then this is the solution:</p>

<pre><code>[FEN "1B6/8/1k6/R7/1n6/2K5/r6p/q1r1n3 - - - 0 1"]
</code></pre>

<p>and Black's last move was b2xc1R++.</p>

<p><sub>I could write an explanation here about I came to this result, but it would essentially be copying @Maxwell's answer. I didn't use it, though.</sub></p>
