 Solution

One explicit future itinerary is obtained by concatenating all binary words, ordered first by length and then lexicographically:Extend it arbitrarily to negative indices, for example by zeros. Every finite binary block occurs in its forward itinerary, and indeed occurs arbitrarily far out as a subword of longer words. Its shift orbit is therefore dense in the full two-symbol shift space. It is not eventually periodic: arbitrarily long runs of zeros and ones occur, together with all other finite words.

The full shift has sensitive dependence because sequences agreeing on an arbitrarily large initial window can differ just beyond it, and shifting brings that disagreement into view. Thus this itinerary supplies a dense aperiodic orbit in the chaotic symbolic system assumed by the question. A merely nonrepeating sequence would not be enough: restricted aperiodic sequences can have much simpler, nonchaotic orbit closures. Nor does symbolic description alone assign a numerical Lyapunov exponent without information about the underlying map and its coding.