In the following section a definition is given followed by some claims.
Is the theory valid?
My work
I am interested in the foundations of math and have thought about concepts like Dedekind-infinite set. In the first paragraph of the wikipedia article on the subject you'll find the sentence
Proposed by Dedekind in 1888, Dedekind-infiniteness was the first definition of "infinite" that did not rely on the definition of the natural numbers.
The definitions/theory below also does not rely on the construction of the natural numbers.
Also, if the ideas are sound and there are already extant expositions of the theory, please provide some references.
Let the function $f: X \to X$ be a given (set) endomorphsim defined on the set $X$.
If $x \in X$ there is a minimal set $\tau^f_x(X) \subset X$ satisfying the following two conditions,
$\tag 1 x \in \tau^f_x(X) $
$\tag 2 \displaystyle \text{The restriction, } f^{\tau}_x \text{, of } f \text{ to } \tau^f_x(X) \text{ defines an endomorphsim } f^{\tau}_x:\tau^f_x(X) \to \tau^f_x(X)$
A set $X$ is said to be $\text{cc-cyclic}$ if there exists a function $f: X \to X$ satisfying
$\quad \forall \, x \in X, \; f^{\tau}_x = f$
The function $f$ is then said to be a complete closed-chain cycle for $X$.
Puzzle Spoiler: If this theory is correct a well known six letter adjective can also be used to describe the $\text{cc-cyclic}$ set $X$.
Claim 1: Induction can be performed on a $\text{cc-cyclic}$ set $X$; here you can start the base case at any element $x_0 \in X$.
Claim 2: The recursion theorem construction technique can be applied (with a simple adaption) to a $\text{cc-cyclic}$ set $X$; here you can begin the functional recursion at any element $x_0 \in X$.
Claim 3: A function that is a complete closed-chain cycle for a set is also a bijection.
Claim 4: Every subset of $\text{cc-cyclic}$ set is also a $\text{cc-cyclic}$ set.
Best Answer
Okay, with your edits today, it makes more sense.
$\tau_x^f$ is sometimes called the orbit of $x$ under $f$ (the "$(X)$" part of the notation is redundant, since $X$ is the domain and codomain of $f$). The condition $f_x^\tau = f$ implies $\tau_x^f = X$.
With that recognition, the inductive principle can be easily proven. Let $Q = \{x \in X\mid P(x)\text{ is true}\}$. Then $x_0 \in Q$ and by the induction hypothesis $f(Q) \subset Q$. Ergo, $\tau_{x_0}^f \subset Q$ by its definition. But since $\tau_{x_0}^f = X$ that gives $Q = X$, or equivalently, for all $x \in X, P(x)$ is true.