Note that any solution will be of the form  $S=\{0,1,s_2,\cdots,s_j,n-1,n\}$ and then $S'=\{0,1,n-s_j,\cdots,n-s_2,n-s_1,n-1,n\}$ is also a solution.

In many of the solutions given by Rob Pratt the solution can remain valid with some shifting of entries to achieve symmetry or near symmetry. For example, if the first entries are $0,1,\cdots,d-1$ and the last $n-(d-1),\cdots n,$ then each of the central entries must be no more that $d$ from their neighbors. 

Here are alternate solutions to those of Rob Pratt which also achieve $m(n).$ I use boldface to indicate that the listed solution is also symmetric.

- $n=7$ $\{0, 1, 2, 5, 6, 7\}$ 
- $n=9$ $\{0, 1, 2, 5, 7,8,9\}$ 
- $\mathbf{n=10}$ $\{0, 1, 2, 5, 8, 9, 10\}$
- $\mathbf{n=11}$ $\{0, 1, 2, 4, 7, 9, 10, 11\}$
- $n=12$  $\{0, 1, 2, 3, 5, 7, 10, 11, 12\}$
- $\mathbf{n=14}$ $\{0, 1, 2, 4, 7, 10, 12, 13, 14\}  $
- $n=17$ $\{0, 1, 2, 5, 8, 9, 12, 15, 16, 17\}$


For $n=2d^2-2$ there are optimal symmetric solutions with $|S|=4d-3$. These are the ones listed by Rob Pratt for $n=6,16.$ 

- $\{0,1,2,3,7,11,15,19,23,27,28,29,30\}$ works for $d=4.$
- $\{0,1,2,3,4,9,14,19,24,29,34,39,44,45,46,47,48\}$ works for $d=5.$

This is, of course, just the scheme suggested by Gerry Myerson with a very minor adjustment.