Idempotency and Fixed-point combinators

A function $f \colon S \to S$ is idempotent if $f \circ f = f$, that is, if for every $x \in S$ one has $f(f(x)) = f(x)$.

A function $f \colon S \to S$ has a fix point if there is some $x \in S$ such that $f(x) = x$ (and such a $x$ is called a fix point of $f$).

Therefore, idempotence is a global property of a function (it says something about all the elements in its domain), while having a fix point is a local property of a function (it claims the existence of some specific element in its domain).

The fact that a function has a fix point or even infinitely many fix points, does not imply that the function is idempotent. Take for instance the function $x \mapsto x\sin x$, whose fix points are $\{\frac{\pi}{2} + 2k\pi \mid k \in \mathbb{Z}\}$, but clearly it is not idempotent.

The same holds for the $\lambda$-calculus. From the fact that every $\lambda$-term $f$ has a fix point (which is $Pf$, where $P$ is a fixpoint combinator), or even infinitely many fix points (it can be proved), it does not follow that $f$ is idempotent. Indeed, it would imply that $\lambda$-terms can only represent idempotent functions. But it is well known that the $\lambda$-calculus is Turing-complete and hence it can represent all recursive functions, including the non-idempotent ones.