Following the definition of an algebraic structure as a set with a collection of $n$-ary operations (https://en.wikipedia.org/wiki/Algebraic_structure) we must consider a set with a permutation (as a unary operation) as an algebraic structure.
The structure is much simpler then a set with a binary operation, so we could expect multiple properties to be defined on the structure first and then applied to more complex structures.
An example of such properties is the decomposition of a finite permutation onto disjoint cycles that becomes the decomposition onto cyclic subgroups of an abelian group.
But instead, we have disconnected terminologies developed for permutations and for magmas/semigroups/groups independently.
One of the most prominent disconnections is the difference between the notions of a cyclic permutation and a cyclic group (Power of an element of a permutation).
Is this a result of a lack of a definition of an algebraic structure of a permutation?
Is there a name for such a structure?
As far as I am aware, there is no name for such a structure, so I will simply call it a permutation. I think there is little use in group theory for allowing infinite permutations to be cyclic, whereas allowing the infinite group $\mathbb{Z}$ to be cyclic is useful for the classification of finitely generated abelian groups. Nonetheless, one could certainly extend the definition of a cyclic permutation to be a permutation $\sigma:X\to X$ such that there exists $x_0\in X$ such that for all $x\in X$ either there exists $n\in\mathbb{Z}$ such that $\sigma^n(x_0)=x$, or $\sigma(x)=x$ (i.e. there is at most one orbit which has more than one element). The "or $\sigma(x)=x$" could be removed to better match the definition of a cyclic group as something which is generated by one element, but the traditional definition of cyclic permutation requires this caveat. I will call $\sigma$ primitive cyclic if the "or $\sigma(x)$" condition can be dropped (i.e. there is exactly one orbit).
In the class of permutations, appending two disjoint permutations together corresponds to a coproduct (i.e. disjoint union) in the category of these algebraic objects. As far as I am aware, there is not too much we can say about this category since permutations are a fairly simple algebraic object, but we do get one theorem analogous to the fundamental theorem of finitely generated abelian groups. Namely, any finitely generated permutation (or even non-finitely generated permutations for that matter) can be decomposed into a coproduct of primitive cyclic permutations.
$\textit{Proof}.$ Let $\sigma:X\to X$ be a permutation. We can define an equivalence relation by declaring $x\sim y$ iff there exists $n\in\mathbb{Z}$ such that $y=\sigma^n(x)$ (i.e. the equivalence class of $x$ is the subpermutation of $(X,\sigma)$ generated by $x$). For each equivalence class, choose a representative and denote the collection of these representatives by $\{x_\alpha\}_{\alpha\in I}$. I claim
$$(X,\sigma)\cong\coprod_{\alpha\in I}\langle x_\alpha\rangle$$
where $\langle x_\alpha\rangle$ denotes the substructure of $(X,\sigma)$ generated by $x_\alpha$ which is clearly a primitive cyclic permutation. Let $\iota_\alpha:\langle x_\alpha\rangle\to X$ be the inclusion maps (which are clearly homomorphisms between these algebraic structures), let $\tau:Y\to Y$ be a permutation, and let $f_\alpha:\langle x_\alpha\rangle\to Y$ be a collection of homomorphisms (i.e. functions such that $\tau f_\alpha=f_\alpha\sigma\vert_{\langle x_\alpha\rangle}$). Then we can define a function $f:X\to Y$ as such: Given $x\in X$, there exists a unique $\alpha\in I$ such that $x\in\langle x_\alpha\rangle$, so take $f(x):=f_\alpha(x)$. By uniqueness of $\alpha$, this is well-defined, and because $\tau f_\alpha=f_\alpha\sigma\vert_{\langle x_\alpha\rangle}$ for all $\alpha$, we see $\tau f=f\sigma$ (i.e. $f$ is a homomorphism). Furthermore, we clearly see $f_\alpha=f\iota_\alpha$ for every $\alpha$. Thus, $(X,\sigma)$ satisfies the universal property of the coproduct, so there is a canonical isomorphism between $(X,\sigma)$ and $\coprod_\alpha\langle x_\alpha\rangle$.