Disclaimer: This is, so far, one of my most downvoted answers on the site. Needless to say, it is perfectly correct, and it answers the question as formulated at the time. I guess one should consider such erratic downvotes as an inherent part of the math.SE experience... Happy reading!
This is to explain how to show the result with no Fourier analysis, using only infinite sequences of heads and tails, so familiar to probabilists, and to draw a parallel with the perhaps better known doubling map.
First, the doubling map. As is well known, the unit interval $[0,1]$ is bimeasurably equivalent, up to null sets for the Lebesgue measure, to the infinite dimensional discrete cube $\{0,1\}^\mathbb N$ by the dyadic mapping $x\mapsto\epsilon=(\epsilon_n)_{n\geqslant1}$ defined, up to null sets for the Lebesgue measure, by
$$
x=\sum_{n=1}^\infty\frac{\epsilon_n}{2^n}.
$$
Then the doubling map $B:[0,1]\to[0,1]$ defined by $B(x)=2x\bmod{1}$, corresponds to the shift $\beta$ defined on $\{0,1\}^\mathbb N$ by
$$
(\beta\epsilon)_n=\epsilon_{n+1}
$$
for every $n$. Thus, for every nonnegative $k$, $$
(\beta^k\epsilon)_n=\epsilon_{n+k}
$$
for every $n$, $B$ leaves invariant the Lebesgue measure on $[0,1]$ and $\beta$ leaves invariant the product uniform measure on $\{0,1\}^\mathbb N$. Furthermore, if $x$ is uniform on $[0,1]$ then $(\epsilon_n)_{n\geqslant1}$ is i.i.d. Bernoulli uniform on $\{0,1\}$ hence, by Kolmogorov zero-one law, the tail sigma-algebra
$$
\bigcap_{k\geqslant0}\sigma(\beta^k)=\bigcap_{k\geqslant0}\sigma(\epsilon_n;n\geqslant k)
$$
is trivial, that is, $\beta$ is ergodic for the uniform measure on $\{0,1\}^\mathbb N$ hence, equivalently, $B$ is ergodic for the Lebesgue measure on $[0,1]$.
Now, to the tent map $T:[0,1]\to[0,1]$ defined by $T(x)=\min(2x,2-2x)$. This time, we encode every $x$ in $[0,1]$ by some sequence $\eta=(\eta_n)_{n\geqslant1}$ in $\{-1,1\}^\mathbb N$ (the replacement of $\{0,1\}$ by $\{-1,1\}$ being merely cosmetic) through the identity
$$
x=\frac12-\frac12\sum_{n=1}^\infty\frac{\eta_n}{2^n}.
$$
To get some familiarity with this somewhat non standard expansion, note that the intervals $(0,\frac12)$, $(\frac12,1)$, $(0,\frac14)$, $(\frac14,\frac12)$, $(\frac12,\frac34)$ and $(\frac34,1)$ correspond to $\eta_1=1$, to $\eta_1=-1$, to $\eta_1=\eta_2=1$, to $\eta_1=-\eta_2=1$, to $\eta_1=-\eta_2=-1$ and to $\eta_1=\eta_2=-1$ respectively.
A key fact is that, if $x$ is uniform on $[0,1]$ then $(\eta_n)_{n\geqslant1}$ is (again) i.i.d. Bernoulli uniform (but this time) on $\{-1,1\}$. Furthermore, the map $T$ corresponds to a twisted shift $\vartheta$ defined by $$(\vartheta\eta)_n=\eta_1\eta_{n+1}$$ for every $n$. The proof is direct: $x<\frac12$ corresponds to $\eta_1=1$, then $$
x=\frac14-\frac12\sum_{n=2}^\infty\frac{\eta_n}{2^n}
$$
hence
$$
Tx=2x=\frac12-\frac12\sum_{n=1}^\infty\frac{\eta_{n+1}}{2^n}=\frac12-\frac12\sum_{n=1}^\infty\frac{\eta_1\eta_{n+1}}{2^n}
$$
Likewise, $x>\frac12$ corresponds to $\eta_1=-1$, then
$$
x=\frac34-\frac12\sum_{n=2}^\infty\frac{\eta_n}{2^n}
$$
hence
$$
Tx=2-2x=\frac12+\frac12\sum_{n=1}^\infty\frac{\eta_{n+1}}{2^n}=\frac12-\frac12\sum_{n=1}^\infty\frac{\eta_1\eta_{n+1}}{2^n}
$$
Now, iterating the shift $\vartheta$ is simple, since, for every nonnegative $k$, $$(\vartheta^k\eta)_n=\eta_k\eta_{n+k}$$ for every $n$, thus, $$
\bigcap_{k\geqslant0}\sigma(\vartheta^k)\subseteq\bigcap_{k\geqslant0}\sigma(\eta_n;n\geqslant k)
$$
is trivial (and the inclusion is an identity), that is, $\vartheta$ is ergodic for the uniform measure on $\{-1,1\}^\mathbb N$ hence, equivalently, $T$ is ergodic for the Lebesgue measure on $[0,1]$.
Best Answer
There are $2^n$ equations for $T^n(x)$, and so there are $2^n$ fixed points. However, some of the fixed points are in the same cycle. So for $T^3(x)$, the fixed points are $\left\{0,\frac29,\frac27,\frac49,\frac47,\frac69=\frac23,\frac67,\frac89\right\}$, and the cycles for $T^3(x)$ are $\left\{\left\{0\right\},\left\{\frac29,\frac49,\frac89\right\},\left\{\frac27,\frac47,\frac67\right\},\left\{\frac69=\frac23\right\}\right\}$. So there are 4 period-3 orbits for $T(x)$, but only 2 with a prime period of 3, and thus only 2 3-cycles.
**EDIT**:
Prove that the tent map has exactly nine 6-cycles. $$T(x)=\left\{\begin{matrix}2x \qquad \ \ \ \ \ 0\le x\le\frac12 \\ 2-2x \quad \frac12< x\le1\end{matrix}\right.$$
$T^n(x)$ has $2^n$ piecewise functions, $n \in N$ by the definition of an iterating function and by $T(x)$.
$T^n(x)$ has $2^n$ fixed points where $T^n(x)=x$. By (1).
Iff $T^n(x)=x$, then $x$ is an element of a period-$n$ orbit of $T(x)$. By definition of orbit.
The fixed points of $T^n(x)$ comprise all elements of the period-$n$ orbits of $T(x)$. By (3).
Given $m<n$ where $m,n \in N$, iff $\frac mn \in N$, then $m$ is a factor of $n$. The factors of $n$ are $\{m_1,m_2,...,m_k\}$ where $k \in N$. By the definition of factor.
$T^n(x)=T^{m_i}(T^{m_j}(x))$, where $m_i \cdot m_j=n$ and $i,j \in N$. By iterating function.
The fixed points of $T^{m_k}(x)$ are also fixed points of $T^n(x)$. By (6).
Given $s$ unique fixed points of $T^{m_k}(x)$, there are $2^n-s$ fixed points of $T^n(x)$ which are not fixed points of any $T^{m_k}(x)$. By (7) and (2).
The fixed points of $T^n(x)$ which are not fixed points of $T^{m_k}(x)$ are all the elements of period-$n$ orbits of $T(x)$ with prime period $n$. By (7) and (3).
There are $n$ unique elements to each $n$-cycle of $T(x)$. By definition of $n$-cycle.
There are exactly $\frac {2^n-s}{n}$ $n$-cycles of $T(x)$. By (10).
For $T^6(x)$, there are $\frac {2^6-s}{6}$ 6-cycles of $T(x)$. By (11).
$s$=number of unique fixed points of $\{T^1(x),T^2(x),T^3(x)\}=2+2+6=10$. By (8) and (5).
There are $\frac {2^6-10}{6}$ 6-cycles of $T(x)$. $$\frac {2^6-10}{6}=\frac {64-10}{6}=\frac {54}{6}=9$$ By (13),(11),(12).
There are exactly nine 6-cycles of $T(x)$. By (14).
Hope that's more comprehensive and helpful.