If $f^{n_o}$ has a fixed point , then does $f$ also has a fixed point , where $f$ is continuous on $\mathbb R$? In relation to this question , To prove : If $f^n$ has a unique fixed point $b$ then $f(b)=b$ , if $f: \mathbb R \to \mathbb R $ is a continuous function such that for some $n_o \in \mathbb N$ the $n_o$th iterate of $f$ has a fixed point , then is it true  that $f$ has a fixed point ? 
Is it possible that if $b$ is a fixed point of $f^{n_o}$ then $b$ must also be a fixed point of $f$?
 A: Suppose a fixed point $b$ of $f^{n_0}$ is not a fixed point of $f$. Then there is a smallest $k > 1$ such that $f^{k}(b) = b$ (this $k$ must be a divisor of $n_0$, but that's not important here). Then $b, f(b),\dotsc, f^{k-1}(b)$ are distinct points.
Let $f^m(b)$ be the largest among these points.
Since $f^{m-1}(b) < f^m(b)$ and $f^m(b) > f^{m+1}(b)$, the continuous function $g(x) = f(x) - x$ attains a positive value in $f^{m-1}(b)$, and a negative value in $f^m(b)$, hence $g$ has a zero in the interval $\bigl(f^{m-1}(b),f^m(b)\bigr)$, that is, $f$ has a fixed point in that interval.
There are conditions that force $b$ to be a fixed point of $f$, for example monotonicity, or contractiveness.
A: Let $f$ be a function on the set $\{-1,1\}$ given by $f(1) = -1$ and $f(-1)=1$. $f^2$ has two fixed points $f^2(\pm 1) = \pm 1$, but $f$ does not have any.
So this idea does not hold in this simple case.

Otherwise examine $f(x) = -x$ which has only one fixed point at $x=0$. $f^2(x)=x$ and every point is a fixed point. This tells us that if $b$ is a fixed point for $f^2$, then it is not necessarily a fixed point for $f$.
