ABCD x 4 = DCBA

First note in the thousands place, $A$ must be 1 or 2. Then note that since $A$ will also be in the ones place in the product with 4, $A$ must be even and must be 2. It then follows $D=4A=8$ since $8\cdot 4=32$ and 2 is in the ones place as desired.

Then we can see that the numbers $BC$ follow the pattern $4C+3=10x+B$ , where $x$ is any amount carried and $4B+x=C$ . Noting that 4B cannot carry to the thousands place $B\leq 2$ . Looking at $4C+3=10x+B$ , $4C+3$ must be odd, so $B$ must be 1 (since 10x is even). Lastly we can solve the system $4C+3=10x+1$ and $4+x=C$ and find $C=7$

Thus the required answer is 2178.