Here is a problem. There are four proofs that I attempted to do but am not sure are correct. Any insight or thoughts concerning the problem and my solutions are most appreciated.
Let $A$ be nonempty set, and let $f:A \rightarrow A$ be a function. Suppose that $f$ is bijective. For each $n \in \mathbb{N}$, let $f^n$ denote the function $A \rightarrow A$ given by
$$f^n = f \circ \cdots \circ f$$
where $f$ is iterated $n$ times.
The function $f^n$ is the $n$-fold iteration of $f$. We now extend the definition of $f^n$ to all $n \in \mathbb{Z}$. Let $f^0 = i_A$ where $i_A: A \rightarrow A$ is the identity mapping. Because $f$ is bijective, it follows that $f^n$ is bijective. Hence $f^n$ has an inverse. For each $n \in \mathbb{N}$, let $f^{-n} = (f^n)^{-1}$. It can be verified that the following two properties are true for all $a,b \in \mathbb{Z}$:
P1.$f^a \circ f^b = f^{a+b}$
P2.$(f^a)^b = f^{ab}$.
Suppose that for all $x,y,z \in A$ the following three properties hold:
T1. $x = f^n(x)$ for some $n \in \mathbb{Z}$,
T2. if $y = f^n(x)$ for some $n \in \mathbb{Z}$, then $x = f^m(y)$ for some $m \in \mathbb{Z}$, and
T3. if $y = f^n(x)$ for some $n \in \mathbb{Z}$ and $z = f^m(y)$ for some $y \in \mathbb{Z}$, then $z = f^p(x)$ for some $p \in \mathbb{Z}$.
Now let $a \in A$. The orbit of $a$ with respect to $f$, denoted $O_a$, is the set defined by $O_a = \{f^n(a) \, | \, n \in \mathbb{Z}\}$.
Let $x,y \in A$. Prove the following properties hold
If $y = f^m(x)$ for some $m \in \mathbb{Z}$, then $O_x = O_y$.
If $y \neq f^n(x)$ for all $n \in \mathbb{Z}$, then $O_x \cap O_y = \varnothing$
$x \in O_y$ if and only if $y \in O_x$
$A = \bigcup_{x \in A} O_x$
Proof of 1. Suppose $y=f^m(x)$ for some $m \in \mathbb{Z}$. By T2, we have $x = f^p(y)$ for some $p \in \mathbb{Z}$. Consequently,
$$y = f^m(x) = f^m(f^p(y)) = f^{m+p}(y)$$
where the last equality holds by P1. Therefore, $y \in O_y$, so that $O_x \subseteq O_y$.
The converse can be established by a similar argument.
Hence, $O_x = O_y$ only if $y = f^m(x)$.
Proof of 2. Suppose that for all $x, y \in A$ and for all $n \in \mathbb{Z}$, we have $y \neq f^n(x)$. It clearly follows that $y \not\in O_x$. Since this is true for every element in $A$ and $\mathbb{Z}$, it must be that $O_x = \varnothing$. Hence, $O_x \cap O_y = \varnothing$.
Proof of 3. Since $f^n$ has an inverse for all $n \in \mathbb{Z}$, it follows that $y = f^n(x)$ if and only if $x = f^{-n}(y)$ where $f^{-n}$ is the inverse of $f^n$ for each $n$. The biconditional implies that $y \in O_x$ if and only if $x \in O_y$, as desired.
Proof of 4. Let $a \in A$. From T1 we know that there exists an $n \in \mathbb{Z}$ such that $a = f^n(a)$. Hence, $a \in O_a$, which follows that $a \in \bigcup_{x \in A} O_x$, so $A \subseteq \bigcup_{x \in A} O_x$. Now suppose that $a \in \bigcup_{x \in A} O_x$. Then for some $x_0 \in A$, we have $a \in O_{x_0}$. Thus there is an $n \in \mathbb{Z}$ such that $a = f^n(x_0)$. By T2, $x_0 = f^m(a)$ for some $m \in \mathbb{Z}$. Since $f^m$ is defined on $A$ it must be that $a \in A$, and so $\bigcup_{x \in A} O_x \subseteq A$.
Hence $A = \bigcup_{x \in A} O_x$
$\blacksquare$
Your proof of $2$ purports to give $O_x=\varnothing$ but $x\in O_x$ contradicts this. You have shown $y\notin O_x$. You need to show that if $z\in O_x$ then $z\notin O_y$ and if $w\in O_y$ then $w\notin O_x$. So suppose $z=f^n(x)=f^m(y)$ is in both, you ought to be able to derive a contradiction.
For 4, A is the union of its singleton subsets and $x\in O_x$. Also $O_x\subset A$. The first tells you that the union contains $A$, the second that $A$ contains the union.