Group $G=\{f : \mathbb{R} \to \mathbb{R}\ |\ f \text{ is a bijective function}\}$ and elements of order $4$ and $9$ 
This question was asked in my abstract algebra quiz and I was unable to completely solve it.

Question : Let $G=\{f : \mathbb{R} \to \mathbb{R}\  |\ f \text{ is a bijective function}\}$ and $\circ$ be operation of composition of functions , then:
A) $G$ is abelian
B) $(G,\circ)$ has an element of order $9$.
C) there are uncountable elements of order $4$.
I proved A wrong . But I think that B can be right but there is no intution for C .
Can you please give a rigorious argument for B and C .
Thanks!!
 A: (B) Let $f$ be the map such that $f:1\mapsto 2\mapsto 3\mapsto 4\mapsto 5\mapsto 6\mapsto 7\mapsto 8\mapsto 9\mapsto 1$, and for every other $x$, $f:x\mapsto x$. The $f$ has order $9$.
(C) For each $\alpha\in\mathbb{R}$ let
$f_\alpha: \alpha+1\mapsto\alpha+2\mapsto\alpha+3\mapsto\alpha+4\mapsto\alpha+1$ and for every other $x$, $f_\alpha:x\mapsto x$. Then each $f_\alpha$ has order $4$, and they are distinct.
A: The more general formulation of your problem is as follows: let $A$ be an infinite set and let $\Sigma(A)$ be the symmetric group on $A$, in other words the group of all permutations of $A$ with respect to composition (of maps). In your particular case $A=\mathbb{R}$.
As $A$ is infinite, for any $n \in \mathbb{N}^{\times}=\mathbb{N} \setminus \{0\}$ there exists an injection $\varphi \colon [1, n] \to A$ -- where by $[r, s]$ I am referring to the natural interval between $r$ and $s$ -- and you can consider the cyclic permutation $(\varphi(k);)_{1 \leqslant k \leqslant n}$, which will of course be of order $n$. Thus $\Sigma(A)$ contains elements of any finite order.
Furthermore, as it is not difficult to show that that the collection of subsets of $A$ whose cardinality is $n$ has itself cardinality equal to $|A|$, it is easy to argue that the subset of all cyclic permutations of $A$ of order $n$ has cardinality $|A|$. Since in your case $|\mathbb{R}|=2^{\aleph_0}$, you have uncountably many elements of any fixed order $n \geqslant 2$.
This is a claim that requires a bit of method in order to demonstrate, but it can be shown that the total number of elements of order $n \geqslant 2$ in $\Sigma(A)$ is actually equal to the order $|\Sigma(A)|=2^{|A|}$ of the full symmetric group on $A$.
