Your conjecture does not hold in general. 

Indeed, for $m=5$ and all natural $n$ we have 
$$d_{m,n}:=d(J_m^{(n)}, \{i/m\}_{i=1}^{m-1}) =\frac2{mn}. \tag{1}\label{1}$$

To prove this, one can check that 

 1. if $n\equiv0$ or $n\equiv4$ ($\mod5$), then $\min\{|1/5-y|\colon y \in J_5^{(n)}\}=\frac2{5n}$;
 2. if $n\equiv1$ ($\mod5$), then $\min\{|4/5-y|\colon y \in J_5^{(n)}\}=\frac2{5n}$;
 3. if $n\equiv2$ ($\mod5$), then $\min\{|2/5-y|\colon y \in J_5^{(n)}\}=\frac2{5n}$;
 4. if $n\equiv3$ ($\mod5$), then $\min\{|3/5-y|\colon y \in J_5^{(n)}\}=\frac2{5n}$. 

On the other hand, one of course has $d_{5,n}\le\frac2{5n}$ for all natural $n$. $\quad\Box$

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To illustrate \eqref{1}, below is the graph $\{(n,m\, n\, d_{m,n})\colon n=1,\dots,100\}$ for $m=5$: 

[![enter image description here][1]][1] 


  [1]: https://i.sstatic.net/bfTxX.png