[proofplan]
We use the conjugacy $h$ to transfer periodic data between the full shift $\sigma: \Sigma_m \to \Sigma_m$ and the restricted diffeomorphism $f|_\Lambda: \Lambda \to \Lambda$. A word of length $n$ defines a bi-infinite periodic sequence, and the least shift period of that sequence is exactly $n$ precisely when the word is primitive. The conjugacy preserves least periods, while cyclic rotations of the word correspond exactly to changing the starting point on the same periodic orbit. Finally, because an alphabet with at least two symbols contains a primitive word of every positive length, the horseshoe has periodic orbits of every positive least period.
[/proofplan]
[step:Set up the symbolic coding and the notion of primitive words]
Let $A_m := \{1,\dots,m\}$ be the alphabet, and let $\Sigma_m := A_m^{\mathbb{Z}}$ be the full two-sided shift space. Let $\sigma: \Sigma_m \to \Sigma_m$ be the shift map defined by $(\sigma s)_k = s_{k+1}$ for every $s = (s_k)_{k \in \mathbb{Z}} \in \Sigma_m$ and every $k \in \mathbb{Z}$. By hypothesis, there is a homeomorphism $h: \Sigma_m \to \Lambda$ satisfying
\begin{align*}
h \circ \sigma = f|_\Lambda \circ h.
\end{align*}
For a word $w = w_0w_1\dots w_{n-1} \in A_m^n$, define the bi-infinite periodic sequence $s_w \in \Sigma_m$ by
\begin{align*}
(s_w)_k = w_r \quad \text{where } r \in \{0,\dots,n-1\} \text{ and } k \equiv r \pmod n.
\end{align*}
We call $w$ primitive if there is no integer $d$ with $1 \leq d < n$ and no word $u \in A_m^d$ such that $w$ is obtained by repeating $u$ exactly $n/d$ times. Equivalently, $w$ is primitive if the sequence $s_w$ has least period $n$ under $\sigma$.
[/step]
[step:Transport symbolic periodic points to periodic points in the horseshoe]
Fix $n \in \mathbb{N}$ and let $w \in A_m^n$ be a primitive word. Since $s_w$ is $n$-periodic under $\sigma$, we have
\begin{align*}
\sigma^n(s_w) = s_w.
\end{align*}
Define the point $x_w \in \Lambda$ by
\begin{align*}
x_w := h(s_w).
\end{align*}
Using the conjugacy relation $h \circ \sigma = f|_\Lambda \circ h$ repeatedly gives
\begin{align*}
(f|_\Lambda)^n(x_w) = (f|_\Lambda)^n(h(s_w)) = h(\sigma^n(s_w)) = h(s_w) = x_w.
\end{align*}
Thus $x_w$ is a periodic point of $f|_\Lambda$ whose period divides $n$.
[guided]
Fix a primitive word $w = w_0w_1\dots w_{n-1} \in A_m^n$. The word determines a symbolic point $s_w \in \Sigma_m$ by repeating the block $w$ indefinitely in both directions. Formally, for each integer $k$, choose the unique residue $r \in \{0,\dots,n-1\}$ with $k \equiv r \pmod n$, and set
\begin{align*}
(s_w)_k = w_r.
\end{align*}
This construction makes $s_w$ fixed by $\sigma^n$, because shifting the sequence by $n$ positions returns every coordinate to the same symbol:
\begin{align*}
\sigma^n(s_w) = s_w.
\end{align*}
Now define $x_w := h(s_w) \in \Lambda$. The conjugacy identity says that applying $f|_\Lambda$ after coding is the same as shifting first and then coding:
\begin{align*}
f|_\Lambda \circ h = h \circ \sigma.
\end{align*}
Iterating this identity $n$ times gives
\begin{align*}
(f|_\Lambda)^n \circ h = h \circ \sigma^n.
\end{align*}
Applying this identity to $s_w$ gives
\begin{align*}
(f|_\Lambda)^n(x_w) = (f|_\Lambda)^n(h(s_w)) = h(\sigma^n(s_w)) = h(s_w) = x_w.
\end{align*}
So the symbolic periodic sequence has been transported to a genuine periodic point in the horseshoe.
[/guided]
[/step]
[step:Use primitivity to identify the least period]
We prove that $x_w$ has least period $n$. Suppose that $(f|_\Lambda)^d(x_w) = x_w$ for some integer $d$ with $1 \leq d \leq n$. Using the conjugacy relation and injectivity of $h$, we obtain
\begin{align*}
h(\sigma^d(s_w)) = (f|_\Lambda)^d(h(s_w)) = h(s_w),
\end{align*}
hence $\sigma^d(s_w) = s_w$. Since $w$ is primitive, the least period of $s_w$ under $\sigma$ is $n$. Therefore $n$ divides $d$. Because $1 \leq d \leq n$, this forces $d = n$. Hence $x_w$ has least period $n$.
[/step]
[step:Quotient primitive words by cyclic rotation to obtain orbits rather than marked points]
Let $w = w_0w_1\dots w_{n-1} \in A_m^n$ be primitive. For each $j \in \{0,\dots,n-1\}$, define the $j$-th cyclic rotation $R_jw \in A_m^n$ by
\begin{align*}
R_jw = w_jw_{j+1}\dots w_{n-1}w_0\dots w_{j-1},
\end{align*}
with indices read modulo $n$. By construction,
\begin{align*}
s_{R_jw} = \sigma^j(s_w).
\end{align*}
Therefore
\begin{align*}
x_{R_jw} = h(s_{R_jw}) = h(\sigma^j(s_w)) = (f|_\Lambda)^j(h(s_w)) = (f|_\Lambda)^j(x_w).
\end{align*}
Thus cyclic rotations of $w$ give exactly the different choices of starting point on the same period-$n$ orbit.
Conversely, let $x \in \Lambda$ have least period $n$ under $f|_\Lambda$, and define $s := h^{-1}(x) \in \Sigma_m$. Since $h^{-1} \circ f|_\Lambda = \sigma \circ h^{-1}$, the sequence $s$ has least period $n$ under $\sigma$. Let $w := s_0s_1\dots s_{n-1} \in A_m^n$. Then $w$ is primitive, $s = s_w$, and the orbit of $x$ is represented by the cyclic rotation class of $w$. This gives the claimed bijection between period-$n$ orbits of $f|_\Lambda$ and primitive words of length $n$ modulo cyclic permutation.
[/step]
[step:Exhibit primitive words of every positive length]
It remains to show existence for every $n \in \mathbb{N}$. Since $m \geq 2$, choose two distinct symbols $a,b \in A_m$. For $n = 1$, the one-letter word $a$ is primitive. For $n \geq 2$, define
\begin{align*}
w_n := \underbrace{aa\dots a}_{n-1 \text{ copies}}b \in A_m^n.
\end{align*}
If $w_n$ were a repetition of a shorter word of length $d < n$, then the number of occurrences of the symbol $b$ in $w_n$ would be a multiple of $n/d$. Since $d < n$, we have $n/d \geq 2$ whenever $d$ divides $n$. But $w_n$ contains exactly one occurrence of $b$, a contradiction. Hence $w_n$ is primitive.
By the preceding steps, each primitive word $w_n$ determines a periodic orbit of $f|_\Lambda$ with least period $n$. Therefore $f|_\Lambda$ has periodic orbits of every positive least period, and those of least period $n$ are precisely the primitive words of length $n$ modulo cyclic permutation.
[/step]
