Answer
$n=4k$, where $k=1,2,3,4,5,...$
Therefore $n=4,8,12,16,...$
Work Step by Step
$f(x)=sin x$
The derivatives:
$f^{(1)}(x)=cos x$
$f^{(2)}(x)=- sin x$
$f^{(3)}(x)=- cos x $
$f^{(4)}(x)=sin x$
At the 4th derivative we return to the original function, then the pattern will repeat, the 5th derivative will be the same as the 1st, and the 6th will be like the 2nd, and so on...
Each four derivative, the pattern will repeat, so we want $sin x$, which is the original function, if we want all the derivatives with the same result, we just have to choose the multiples of 4..
If $n=4k$, where $k=1,2,3,4,5,...$
Then $f^{(n)}(x)=sin x$