We solve the generalized case where $a, b$ may be irrational. By Kronecker's this only holds if $a$ is some rational multiple of $b$. If $a = qb$ for $q \ne$ then this implies that $\{b\}$ is not equidistributed mod denom of $q$ so $a$ and $b$ must both be rational. Note that if $\{na\}$ is $0$ then $\{nb\}$ must be as well so the denominator of $b$ divides $a$. If the denominators aren't equal, then increasing $n$ by $b$ amounts to shifting $\{na\}$ by some nonzero constant. This forces $\{nb\} \le 1/2$ for all $b$, which is impossible. We can now reduce the problem to be mod the common denominator. Take $n$ which is inverse of the numerator of $a$. It then implies the same for the numerator of $b$, so they must be equal.

Write $a=\frac{p}{q}$ where $(p,q) = 1$. Then $\{qa\} \geq \{qb\}$ implies $\{qb\} = 0$, so $qb$ is a positive integer, i.e. $b=\frac{r}{q}$. So now $\left\{\frac{np}{q}\right\} \geq \left\{\frac{nr}{q}\right\}$ for all $n$, as well as $p<q$ and $r<q$ (since the problem condition requires $a,b \in (0,1)$) and aim to show $p=r$. Note that $n=1$ insists on $p \pmod q \geq r \pmod q$, while $n=q-1$ insists the opposite, thus equality must hold and we are done.

$a=\frac{p}{q}$ $gcd(p;q)=1$. For $n=q$, we have $0\geq \{qb\}\Rightarrow b=\frac{m}{q}$. If $m>p$, then $n=1$ gives $\{a\}<\{b\}$ If $m<p$, then $n=q-1$ gives $$\{na\}=\{p-\frac{p}{q}\}=1-\frac{p}{q}$$$$\{nb\}=\{m-\frac{m}{q}\}=1-\frac{m}{q}$$$$1-\frac{p}{q}<1-\frac{m}{q}\Rightarrow \{na\}<\{nb\}$$So only case left is $m=q$, which gives $a=b$,which is an obvious answer