[NB] This argument as stated is incorrect, and the error is pointed out by Yuval in the comments below. One can indeed use the De Bruijn graph to get all permutations, exactly as I described below, but as Yuval's comment implies, one will also get plenty of non-permutations. The length of the string $S$ is not what my argument says it is, and the problem of the MINIMAL LENGTH sequence seems to require a different argument.

## This is very dangerous because all those undeserved points might cause someone to dismiss what might be an interesting and difficult question.

Since there are $n!$ permutations of $n$ symbols, the string $S$ must have length
at least $n!+n-1$. That is, there must be $n!$ starting positions for a permutation, and the last starting position must be succeeded by $n-1$ symbols.

There is in fact such a string, which is obtained from an Eulerian path in the De Bruijn graph $B(n,n)$. This is a directed graph whose vertices are all strings of $n-1$ distinct symbols, with a directed edge from $u$ to $v$ if there is a symbol $s$ such that $v$ has the form $u_1s$, where $u_1$ is $u$ with the left-most symbol deleted. We label the edge with $s$. Every vertex has the same in-degree and out degree so the graph has an eulerian path. The labels of the edges on any such path, together with the string where the path ends, gives a minimal sequence of the kind you want. See this Wikipedia article http://en.wikipedia.org/wiki/De_Bruijn_sequence for terminology, references and a nice explanation.