Let $f$ be a permutation on a set $U$. Suppose that for every $x\in U$, the sequence $$x,f(x),f\circ f(x), f\circ f\circ f(x), \ldots$$ stabilizes, and let this value of $x$ be called $f^\infty(x)$. Suppose that $f=f^\infty$. Then, $f$ is the identity function on $U$.
Here is my take on it. Let $f:U\to U$ be bijective, hence, a permutation on $U$. Then, because $f=f\circ f\circ f\cdots$, then $f\circ f= f\circ f \circ f \circ\cdots= f$, so $f\circ f=f$, and because it is bijective we can take its inverse giving us $f^{-1}\circ f\circ f=f^{-1}\circ f\iff f=I$. Could anyone confirm this? Thanks.
Let me try to rewrite the question and please tell me if this is what the question means. If this is not the case, Ill delete the answer.
If this is what the question means, then you are kind of correct. This is because $f^\infty=f\circ f^\infty$. If $f=f^\infty$, then $f=f\circ f^\infty$. Since $f$ is a bijection, $f^\infty$ must be the identity function on $U$. But as $f=f^\infty$, $f$ is also the identity function on $U$.
However, if $U$ is equipped with a topology so that it is possible to discuss limits, then this is also a valid interpretation of the problem. The proof remains the same. My first interpretation is actually a consequence of the second interpretation. That is, in the first interpretation, $U$ is equipped with the discrete topology.