= Solution
The set $I^n$ consists precisely of points whose first $n$ positions $x,f(x),\ldots,f^{n-1}(x)$ all lie in $L\cup R$. Hence
$$
\Lambda=\{x\in[0,1]: f^k(x)\in L\cup R\text{ for every }k\geq0\}.
$$
Deleting the first position preserves this condition, proving $f(\Lambda)\subseteq\Lambda$. For the reverse inclusion, if $y\in\Lambda$, take $x=g_L(y)$. Then $x\in L$ and $f^k(x)=f^{k-1}(y)\in L\cup R$ for every $k\geq1$. Thus $x\in\Lambda$ and $f(x)=y$. In fact both <inverse branches> give preimages in $\Lambda$. Therefore
$$
\boxed{f(\Lambda)=\Lambda,}
$$
which proves full invariance, not merely forward invariance.
For any infinite symbolic <sequence> $s$, the <itinerary cylinders> $I_{s_1\cdots s_n}$ are nonempty compact <nested intervals>. The <nested interval theorem> gives at least one point in their intersection. Every $x\in\Lambda$ has a unique itinerary because $L\cap R=\varnothing$. To justify that the itinerary also determines a unique point, it suffices to show the <interval> lengths shrink. The fact that <logistic inverse branches contract in an angular coordinate> works for every $\mu>4$; a direct claim that $|g_L'|,|g_R'|<1$ in the original coordinate would fail for some $\mu$ just above four.
Use the <homeomorphism> $h(u)=\sin^2(\pi u/2)$ of $[0,1]$. The transformed <inverse branches> are
$$
G_L(u)=h^{-1}(g_L(h(u)))=\frac1\pi\arcsin\left(\frac2{\sqrt\mu}\sin\frac{\pi u}{2}\right),\qquad
G_R(u)=1-G_L(u).
$$
Their <derivatives> satisfy
$$
|G_L'(u)|=|G_R'(u)|=
\frac{\cos(\pi u/2)}{\sqrt\mu\sqrt{1-(4/\mu)\sin^2(\pi u/2)}}\leq\frac1{\sqrt\mu}<1.
$$
The inequality follows by squaring and using $4/\mu<1$. Thus every length-$n$ transformed <itinerary cylinder> has length at most $\mu^{-n/2}$. Since $|h'|\leq\pi/2$, its original <interval> has length at most $(\pi/2)\mu^{-n/2}$ and shrinks to a point. This gives the claimed one-to-one identification with the one-sided <full shift>.
Finally part (ii) gives
$$
f(x)\in\bigcap_{n\geq2} I_{s_2\cdots s_n},\qquad
\boxed{s_1s_2s_3\cdots\ \longmapsto\ s_2s_3\cdots.}
$$
Thus $f$ acts by the <left shift>. The identification is a <topological conjugacy>: matching long prefixes forces closeness by the shrinking estimate, while distinct finite <itinerary cylinders> are disjoint compact sets and so their inverse coding is <continuous>. In particular $f$ need not be one-to-one on $\Lambda$, just as a one-sided <left shift> has two possible first symbols before any given tail.
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