In general, if we have a function $f_1:A\to B$ and a function $f_2:B\to A$ such that $$f_1 \circ f_2=\operatorname{id}$$ and $$f_2 \circ f_1=\operatorname{id}$$ then we would call $f_1$ the inverse function of $f_2$, or vice versa. If a function is its own inverse, we call it an involutory function.
My question is this: if we have some functions $f_1,f_2,...,f_{n-1},f_n$ $$f_1:A_1\to A_2$$ $$f_2:A_2\to A_3$$ $$...$$ $$f_{n-1}:A_{n-1}\to A_n$$ $$f_n:A_n\to A_1$$ such that their composition results in $\operatorname{id}$ no matter in what order they are composed, what do we call them?
Furthermore, if a function results in $\operatorname{id}$ when composed with itself $n$ times, what do we call it? I've been calling such a function "n-involutory"... is there already a name for this?
$\DeclareMathOperator{\Id}{id}$Let $n \geq 3$. A set of functions $f_{1}, \dots, f_{n}$ as in your first question might be termed a "commuting family of $n$ permutations (bijections) of a set $A$":
In order for the compositions to be the identity no matter what order the functions are composed, they all must be permutations of a single set $A$. For instance, taking $n = 3$ we have $$ f_{1}f_{2}f_{3} = f_{1}f_{3}f_{2} = \Id_{A_{1}}, $$ so $f_{2}f_{3} = f_{3}f_{2} = f_{1}^{-1}$ is bijective, which implies in particular that $A_{2} = A_{3}$. Permutation of indices shows $A_{1} = A_{2}$ as well. A similar argument handles more than $3$ functions.
That is, each $f_{i}$ is a permutation of the single set $A = A_{1} = \cdots = A_{n}$.
Any two of these maps commute: For instance, $f_{1}f_{2}$ and $f_{2}f_{1}$ are both the inverse of $f_{3} \cdots f_{n}$.
For your second question, a mapping $f:A \to A$ whose $n$-fold self-composition is the identity (and whose $k$-fold composition is not the identity if $0 < k < n$) would naturally be called a permutation of order $n$, "order" being used in the sense of group theory.
The orbits of such an $f$ are obviously finite (with cardinality dividing $n$), and $f$ acts by cyclic permutation on each orbit. This is, $f$ is a product of disjoint cycles, each of length dividing $n$ and at least one of length $n$.