Suppose a sequence $s_k = (a_1, a_2, \ldots)$ has $s_k = s_{k+1}$. Then $a_i = a_{a_i}$, and by injectivity of $f$ (and hence interations of $f$), we have $a_i = i$. So there is just one fixed sequence.

Here are a family of sequences where $s_1 = s_{2n + 1}$. Only the first $2^n + 1$ terms are permuted, and the rest of the sequence is a copy of the identity. If any number $a_i = i$ at some point during the iteration of sequences, then it will remain stationary forever, so we need only consider the first $2^n + 1$ terms. $$ s_1 = (2^n + 1, 1, 2, 3, \ldots, 2^n) $$ $$ s_2 = (2^n, 2^n + 1, 1, \ldots, 2^n - 1) $$ $$ s_3 = (2^n - 2, 2^n - 1, 2^n, 2^n + 1, 1, \ldots, 2^n - 3) $$ $$ s_4 = (2^n - 6, 2^n - 5, 2^n - 4 , 2^n - 3 , 2^n - 2, 2^n - 1, 2^n, 2^n + 1, 1, \ldots, 2^n - 7)$$ $$ \cdots $$ $$ s_{n+1} = (2, 3, \ldots, 2^n + 1, 1) $$ $$ s_{n+2} = (3, \ldots, 2^n + 1, 1, 2) $$ $$ s_{n+3} = (5, \ldots, 2^n + 1, 1, 2, 3, 4) $$ $$ \cdots $$ $$ s_{2n+1} = (2^n + 1, 1, 2, 3, \ldots, 2^n) $$

Doing the same thing beginning with $2^n$ instead of $2^n + 1$ leads to the identity. Doing it with $2^n + k$ where $0 < k < 2^n$ leads to a cycle with some term $s = (2^n + 1, 2^n + 2, \ldots, 2^n + k, 1, 2, \ldots, 2^n)$.

If there is some simple bounding condition as follows, then iterations will cycle locally, but may not globally. Suppose there is a sequence of natural numbers $0 = m_0 < m_1 < m_2 < \cdots$ such that for every $j$, $a_i \leq m_j$ for all $i \leq m_j$. Then the first $m_1$ terms will remain in the first $m_1$ places, the next $m_2 - m_1$ terms will remain in the next $m_2 - m_1$ places, etc. Since there are only $(m_{j+1} - m_j)!$ permutations of each of these sets, they must all eventually be cyclic individually. But we can choose these cycles to have no common multiple, e.g. take $m_{i+1} - m_i = 2^i + 1$, and repeat the construction above.

Suppose a sequence $s_k = (a_1, a_2, \ldots)$ has $s_k = s_{k+1}$. Then $a_i = a_{a_i}$, and by injectivity of $f$ (and hence iterations of $f$), we have $a_i = i$. So there is just one fixed sequence.

Here are a family of sequences where $s_1 = s_{2n + 1}$. Only the first $2^n + 1$ terms are permuted, and the rest of the sequence is a copy of the identity. If any number $a_i = i$ at some point during the iteration of sequences, then it will remain stationary forever, so we need only consider the first $2^n + 1$ terms. $$ s_1 = (2^n + 1, 1, 2, 3, \ldots, 2^n) $$ $$ s_2 = (2^n, 2^n + 1, 1, \ldots, 2^n - 1) $$ $$ s_3 = (2^n - 2, 2^n - 1, 2^n, 2^n + 1, 1, \ldots, 2^n - 3) $$ $$ s_4 = (2^n - 6, 2^n - 5, 2^n - 4 , 2^n - 3 , 2^n - 2, 2^n - 1, 2^n, 2^n + 1, 1, \ldots, 2^n - 7)$$ $$ \cdots $$ $$ s_{n+1} = (2, 3, \ldots, 2^n + 1, 1) $$ $$ s_{n+2} = (3, \ldots, 2^n + 1, 1, 2) $$ $$ s_{n+3} = (5, \ldots, 2^n + 1, 1, 2, 3, 4) $$ $$ \cdots $$ $$ s_{2n+1} = (2^n + 1, 1, 2, 3, \ldots, 2^n) $$

Doing the same thing beginning with $2^n$ instead of $2^n + 1$ leads to the identity. Doing it with $2^n + k$ where $0 < k < 2^n$ leads to a cycle with some term $s = (2^n + 1, 2^n + 2, \ldots, 2^n + k, 1, 2, \ldots, 2^n)$.

If there is some simple bounding condition as follows, then iterations will cycle locally, but may not globally. Suppose there is a sequence of natural numbers $0 = m_0 < m_1 < m_2 < \cdots$ such that for every $j$, $a_i \leq m_j$ for all $i \leq m_j$. Then the first $m_1$ terms will remain in the first $m_1$ places, the next $m_2 - m_1$ terms will remain in the next $m_2 - m_1$ places, etc. Since there are only $(m_{j+1} - m_j)!$ permutations of each of these sets, they must all eventually be cyclic individually. But we can choose these cycles to have no common multiple, e.g. take $m_{i+1} - m_i = 2^i + 1$, and repeat the construction above.