A burning question when reading algebra!! Given a bijection g from a set  A into itself and let MAP(A,B) be the set of all mappings from A into a set B.
Consider the equation f○g=f for a f in MAP(A,B).
Trivially, if there exists a partition P of A such that g(S)=S for each S in P, then any f in MAP(A,B) which is constant on S for each S in P solve the equation.
If P is finite of cardinality n and B is so of m, then there are m^n solutions for one such P.
My question is:
(1.) Under what conditions imposed on g as well as A can we assure the existence of such P? What further conditions are needed to ensure uniqueness?
(2.) If in the negation of the preceding conditions, is the equation still solvable? Is there any other way to approach the equation?
 A: Not an answer, but this observation might be helpful.
Let $f:A\to B$ be a function.
On $A$ let $P$ be the partition of fibres of $f$. 
That is: $P=\{f^{-1}(\{b\})\mid b\in\text{im} f\}$.
Then for any function $g:A\to A$ we have: $$f\circ g=f \iff g(S)\subseteq S\text{ for every }S\in P$$
A: Oh! I think I have sorted somethings out when chilled down in contemplation.
Given a Group G and a set S and let ρ(S) be an action induced an evolution ρ.
For an x in G, let f be in MAP(S,T) for a set T such that f○ρ(x)=f,
then f=f○ρ(e)=f○ρ(x*x^(-1))=f○[ρ(x)○ρ(-x)]=[f○ρ(x)]○ρ(-x)=f○ρ(-x)
and so f○ρ(x^n)=f for all integers n.
Let H be the cyclic subgroup of G generated by x, i.e. H = < x >, and let H operate S by ρ(S).
For all s in S, by orbit decomposition of S, there exists an orbit through a t in S, say H.t such that s is in H.t, then s=ρ(x^m)(t) for an integer m and so ρ(x)(s)=ρ(x^(m+1))(t).
Since m is an integer, so is m+1 and so ρ(x)(s) is still in H.t.
Consequently, since f○ρ(x)=f, f is constant on each orbit. Conversely, if there exists an orbit H.t such that f(u)≠f(v) for some u,v in H.t, then one cannot assume that f○ρ(x)=f.
Hence the solution set of the equation is exactly those f in MAP(S,T) such that f is constant on each orbit in S.
Now let G=Z, the additive group of integers, S=Perm(A), the group of bijections from a set A into itself, ρ in MAP(G x S, S) be given by ρ(m, g)=g^(○ m) for a m in Z and a g in Perm(A).
Trivially, ρ(mn, g)=ρ(m, ρ(n, g)) for all integers m and n and ρ(0, g) is the identity map of A. Hence ρ induce an action ρ(Perm(A)) by Z on Perm(A).
For a positive integer m, the solution set of the equation f○g^(○ m)=f for a f in MAP(A,B) is exactly those f in MAP(A,B) such that f is constant on each orbit in Perm(A) when Perm(A) is operated by Z/(m) under the action ρ(Perm(A)).
Lastly, set m=1.
Please comment on this. I would really like to start reading the next chapter on ring structure next weekend ^.^
