Black to play in the starting position

I believe you mean one "tempo", in this case, yes you can arrive at a legal position on the board from the starting position where black is up a tempo.

Yes

No, that is not the starting position.

The answer is no, it is physically impossible to reach the starting position with Black to move.

The reasoning is simple. All that can move in the starting position are pawns and knights. If you move a pawn, it cannot go back. Knights, on the other hand, because of how they move, cannot lose a tempo, unlike a Bishop, Rook, or Queen.

Now any position other than the starting position where White can triangle or have 3 squares for a rook, bishop, or queen, yea it can be done. For example, the one I quoted. The position with a WP on e4 and BP on e5 is possible, even after 1.e4 e5, you can reach that position with Black to move - 1.e4 e5 2.Bc4 Bc5 3.Bd3 Bf8 4.Bf1 with Black to move.

But not the starting position!

Seems like their is confusing.

What does the OP consider the “starting position” to be?

The diagram I showed would be starting position of King Pawn Opening.

Unless, the OP is talking about the initial board set up.

If we are talking about initial board set up, Than I think my answer would still be Yes.

I think players are allowed to change the color and design of their chess pieces so technically a player would use Black Clown Figurines to represent the (White side pieces on board).

I am pretty sure their is a ruling about chess pieces some where.

Your first sentence says it all! You said yourself that yours is the starting position OF KING PAWN OPENING. It is not "The Starting position". The starting position sees no pieces or pawns on the 3rd through 6th ranks.

By that logic, EVERY possible chess position can be considered "the starting position".

@1

"Is it possible to arrive with a sequence of legal moves in the starting position but white loses one time?" ++ No, you cannot win or lose a tempo with knight moves, or with rook moves restricted to 1 square .

I will disagree that you can't lose a tempo with knight moves, I think I've heard that said before but it's plainly not the case since you can use variable amounts of tempo getting to any one square, hence losing tempos.

I think to be more precise - you can't lose or gain an uneven amount of tempo and end up on the same square, some people have called this effect a knight's parity.

The simplest proof of this, and I am just coming up with this as I go here, is that a knight always goes from either a light-squared to a dark-squared one or vice versa, hence it will always take an even number of jumps to go back to its original square. Since you have two knights to begin with, you could swap their squares with the one on g1 ending up on the lightsquared b1 and vice versa and each of them would take an odd amount of moves to get there. However an odd number + an odd number is an even number, since every odd number can be considered an even number with a 1 tacked on at the end - and you could add up the two 1s tacked on at the end to make the number even again. For example 5 + 7 = (4+1) + (6+1) = 4 + 6 + 2, which is of course even. And because black faces the same problem, the amount of moves for him to reach the starting position will also be even - hence it's white's turn again.

Even if white had three knights it also wouldn't work, since a light-square starting knight would have to go to a dark-squared and a dark squared to a light-squared (odd + odd = even), then the remaining knight would just be going back to a dark square from a dark square which we already know will be even. No matter how many knights you have, if you are bringing them back to the squares they started on (even if different knights come back to each square), it's going to take an even number of moves.

Can't be done with today's chess conventions. Shuffling a Rook between two adjacent squares always takes two moves. Empty the board apart from one Knight. Pick two squares. Try to move the Knight from one to another first in an odd number of moves then in an even number of moves. One or other will be impossible (as far as I'm aware). In the Victorian era, of course, Black to move first in the starting position was easy: the "White moves first" convention had not fully entered the rules (Black was considered the more lucky colour, the convention was intended to counteract that perceived advantage).