Answer: 937

Small Hint:

The condition says $N^2 \equiv N \pmod{10^4},$ i.e. $N(N-1) \equiv 0 \pmod{2^4 \cdot 5^4}$

Big Hint:

$N$ and $N-1$ are coprime, so $16$ and $625$ each divide one of them; only one of the four cases has nonzero leading digit $a$

Solution:

The condition is $N^2 \equiv N \pmod{10^4},$ that is, $N(N-1) \equiv 0 \pmod{2^4 \cdot 5^4}.$ Since consecutive integers are coprime, $16$ divides one of $N,$ $N - 1$ and $625$ divides one of them. This gives four cases modulo $10000:$ $N \equiv 0,$ $N \equiv 1,$ $N \equiv 625$ (which is $0$ mod $625$ and $1$ mod $16$), and $N \equiv 9376$ (which is $0$ mod $16$ and $1$ mod $625$).

The last four digits $abcd$ must have $a \ne 0,$ which rules out $0000,$ $0001,$ and $0625.$ So $abcd = 9376$ — for instance $9376^2 = 87909376$ — and $abc = 937.$