= Solution
One explicit future itinerary is obtained by concatenating all binary words, ordered first by length and then lexicographically:
$$
\boxed{0\,1\,00\,01\,10\,11\,000\,001\,010\,011\,100\,101\,110\,111\,\cdots.}
$$
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.
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