Answer: No. See my proof inside.

Let us say (aNb)=k(bNa) where a and b be are digits and N is an (n-1)-digit number. That is,
10^n*a+10N+b = k*(10^n*b+10N+a).
Note a>b.

Then (10^n-k)*a - (k*10^n-1)*b = 10(k-1)N
The key observation is -ka+b=0(mod 10).

Case 1 k=2. b<5.
We have
(10^n-2)*a - (2*10^n-1)*b = 10N.
Then -2a+b=0 (mod 10).
So b is even. Let b=2, a=6. But a is too large.
Let b=4, a is 2 or 7, both of which are too small.
There is no solution for k=2.

Case 2 k=3. b<4.
We have
(10^n-3)*a - (3*10^n-1)*b = 20N.
Then -3a+b=0 (mod 10).
Let b=1, a=7. But a is too large.
Let b=2, a is 4 which is too small.
Let b=3, a is 1 which contradicts a>b.
There is no solution for k=3.

Case 3 k=4. b<3.
We have
(10^n-4)*a - (4*10^n-1)*b = 30N.
Then -4a+b=0 (mod 10).
So b is even. b=2 only Then a=8 (we ignore a=3 Plugging in these we obtain N=-10.
There is no solution for k=4.

For k=5, 6, 7, 8, and 9, b=1. We can easily derive contradictions.

Therefore, there are no two numbers satisfying the given property.