No, of course not. For instance, if $X_1$ has the standard Cauchy distribution, then $\overline X$ also has the standard Cauchy distribution -- see e.g. [here][1]. So, then $s(x)=\frac1n\,t(x)$. 


  [1]: https://en.wikipedia.org/wiki/Cauchy_distribution#Sum_of_Cauchy_distributions