Fixed point and fractional iteration: if $F(k)=k$ then $F^{1\over n}(k)$ is another fixed point of $F$ My knowledge of the fixed points and iteration equals zero, same for the notation and terminology but I really need to know if this deduction has trivial errors or is really as nice as it seems.

I would like to prove the following:
Notation 1 - Given a function $f:A\rightarrow A$ define the set $Fix(f)\subseteq A$ as the set of $f$'s fixed points
$Fix(f):=\{\phi:f(\phi)=\phi\}$
Notation 2 - Given a function $f:A\rightarrow A$ and the definition of function composition $\circ$ define the function $f^n$ by recursion
$i)$ $f^0:=\operatorname{id}_A$
$i)$ $f^{n+1}:=f\circ f^n$
Definition 1 - Given a function $F:X\rightarrow X$, the " $1\over n$-iterate" of $F$ is a function $\Psi:X\rightarrow X$ with this property
$\forall x(x\in X) (\Psi^n(x)=F(x))$
I guess that we can write $\Psi=F^{1\over n}$

To Prove - If $k\in X$ is a fixed point of $F$ and exists a fucntion $\Psi$ such that $\Psi^n=F$ then $F^{1\over n}(k)$ is a fixed point of $F$
$$k\in Fix(F)\implies \forall n(n\ge1)(F^{1\over n}(k)\in  Fix(F))$$

Proof 1 - For a fixed $n\gt 1$ define $\lambda:=\Psi(k)=F^{1\over n}(k)$
$\lambda:=\Psi (k)=\Psi(\Psi^n(k))$ because $k=\Psi^n(k)$
$\lambda=\Psi^n(\Psi(k))$ because iterates commute
$\lambda=\Psi^n(\lambda)$ because $\Psi(k)=\lambda$ by definition
Since $\Psi^n=F$ by definiton we conclude that $\lambda=F(\lambda)$ and thus 
$$\lambda\in Fix(F)$$
Anyways this proof seems weird to me... I feel like there is something missing: I want to prove that for every natural number (greater than zero), if $F^{1\over n}$ exists,  $F^{1\over n}(k)$ is a fixed point so maybe I need to use induction but I really don't know how I could do it

Questions

$1)$ - Is this proof correct? If yes and it is a known result, can you
   add some info about it?
$2)$ - Is it possible to use induction for the proof? Or it is useless?
$3)$ - If the proof is correct, is this a result that can be
   strengthened? In fact it seems to me that the real generalized result
   would be something like $$k\in Fix(F)\implies \forall q(q\in\Bbb
 Q\land 0\lt q \lt 1)(F^{q}(k)\in Fix(F))$$

 A: I do not believe there is a short answer that will adress all your questions.
However consider the analytic solutions and the complex numbers.
Then we might have $f(z_0)=z_0$ where $z_0$ is a complex number and $f$ is an analytic function.
Now consider the function $F(z+1) = f(F(z))$.
Now your question is equivalent to $if$ $F(q)=z_0 $ where $q$ is finite ,
is it Always true that $F(q + {1\over m} + n) = F(q+{1\over m})$ for positive integers $n,m$ ?
And the answer is NO !
( In fact this has been mentioned and investigated by tommy1729 , Gottfried and probably all frequent posters at the tetration forum )
Here is the proof :
$F(q + {1\over m} + n) = F(q+{1\over m})$ implies that $F(q + {a\over m} + an) = F(q+{a\over m})$ for positive integer $a$.
( Notice the subtle $an$ term , not just $n$ ; the $a$ th iterate of $F({1\over m} + n)$ => $F({a\over m} + an)$. You might want to think about this. )
As a consequence since $q$ is finite and $an$ is an integer ... and $f$ is analytic then by analytic continuation $F$ ... and the fact that the rationals are dense in the reals ... $F$ is periodic with period $an$.
But superfunctions tend not to be periodic.
( superfunction means inverse abel function for those unfamiliar with the term )
Notice that $F(z+\theta(z))$ where $\theta$ is a one-periodic function satisfies the same functional equation but the same problem/solution (proof steps above) holds ... unless of course ( the "new" ) $q$ is no longer finite.
I can find you some related threads on the tetration forum if you like.
You know , periodicity is an important concept the re-occurs often.
This is not Always clear from (basic) traditional education apart from trigonometry. 
Guess that gives you some insight why your idea is a bit too optimistic.

(ordinary reader may stop reading , this is a specialized comment )
As for tetration The kneser solution has the primary fixpoints of exp at +/- oo $i$. And there at oo $i$ the function is indeed flat and periodic. Hence there it is true what you assume ... by " design of the superfunction " ,  however I assume the fixpoints have no finite $q$ for the kneser function. 
Formalizing that might be difficult.
I considered thinking about univalent zones , but since exp is a chaotic map that seems difficult.
Maybe that would make a good question at the tetration forum.

A: I think the main thing missing from the proof is a description of functions admitting an $n^{\rm th}$ root.  There is nothing strange (that I can see) in the steps of your proof but it begs the question - for what $F$ is the set $\{ \Psi : \Psi^n = F \}$ nonempty?
Starting with $X = \mathbb{R}^k$, I believe any linear transformation admits a ${1\over n}$-iterate $\Psi$, but that still leaves the field very open.
