Consider rational approximations of $\sqrt{2} + \frac{1}{a}$. By Dirichlet approximation, we can find integers $m, t$ with $1 \leq t < a$ such that \[\left|m - t\left(\sqrt{2} + \frac{1}{a}\right)\right| < \frac{1}{a}\]The above translates to $\frac{t-1}{a} < m-t\sqrt{2} < \frac{t+1}{a}$. In particular, remark that it follows $0 < m-t\sqrt{2} < 1$ because of $t \in [1, a)$. Thus $m-t\sqrt{2} = \{-t\sqrt{2}\}$. In particular, since this number and $t/a$ are less than $1/a$ apart, it follows that the intervals \[ (t/a, (t+1)/a) \text{ , } \left(\{-t\sqrt{2}\}, 1/a + \{-t\sqrt{2}\}\right)\]intersect. Because $\sqrt{2}$ is irrational, the numbers $\{n\sqrt{2}\}$ are equidistributed mod $1$. Since the above intervals lie completely inside $(0, 1)$ (from the earlier bounding) it follows that we can select a positive integer $n$ such that $\{n\sqrt{2}\}$ lies in both of the above intervals. We claim that this $n$ works, which will finish the problem. First, remark that $a\{n\sqrt{2}\}$ lies strictly between $t$ and $t+1$, implying that $f(n) = n+t \ne n$. Now, since \[\{n\sqrt{2}\}+\{t\sqrt{2}\} > \{-t\sqrt{2}\}+\{t\sqrt{2}\} = 1\]we have $\{(n+t)\sqrt{2}\} = \{n\sqrt{2}\}-\{-t\sqrt{2}\} < 1/a$, implying that $\lfloor a\{(n+t)\sqrt{2}\} \rfloor = 0$ or that $f(n+t) = n+t$. Thus $n$ satisfies $f(f(n)) = f(n)$ and $f(n) \ne n$, as desired.

I would be amused if someone solves this problem while proceeding forwards. I had to back construct $n$ by going all the way "back" to considering the real number $\sqrt{2}+1/a$.