Let $f$ be the function from $\mathbb{N}^{*}$ to $\mathbb{N}^{*}$ defined by (for all $n$ in $\mathbb{N}^{*}$):

$$
\begin{aligned}
f(1) & =1 \\
f(3) & =3 \\
f(2 n) & =f(n) \\
f(4 n+1) & =2 f(2 n+1)-f(n) \\
f(4 n+3) & =3 f(2 n+1)-2 f(n)
\end{aligned}
$$
The study of small values and the omnipresence of powers of 2 quickly lead to a more pleasant expression of $f$: $c^{\prime}$ is the function that reverses the binary representation. The equations in the statement clearly define $f$ completely, so it suffices to verify that this other function, which we will call $g$, also satisfies these conditions. Fix $n \in \mathbb{N}^{*}$ and denote ${\overline{b_{1} b_{2} \ldots b_{n}}}^{2}$ as the binary representation of $n$. Then $g(1)=g\left(\overline{1}^{2}\right)=\overline{1}^{2}=1, g(3)=g\left(\overline{11}^{2}\right)=\overline{11}^{2}=3$, and:

$$
g(2 n)=g\left({\overline{b_{1} b_{2} \ldots b_{n} 0}}^{2}\right)={\overline{0 b_{n} \ldots b_{2} b_{1}}}^{2}={\overline{b_{n} \ldots b_{2} b_{1}}}^{2}=g\left({\overline{b_{1} b_{2} \ldots b_{n}}}^{2}\right)=g(n)
$$

We also have:

$$
\begin{aligned}
g(4 n+1) & =g\left({\overline{b_{1} b_{2} \ldots b_{n} 01}}^{2}\right)={\overline{10 b_{n} \ldots b_{2} b_{1}}}^{2}=2^{n}+{\overline{1 b_{n} \ldots b_{2} b_{1}}}^{2} \\
& =2 \cdot{\overline{1 b_{n} \ldots b_{2} b_{1}}}^{2}-{\overline{b_{n} \ldots b_{2} b_{1}}}^{2}=2 g(2 n+1)-g(n)
\end{aligned}
$$

and:

$$
\begin{aligned}
g(4 n+3) & =g\left({\overline{b_{1} b_{2} \ldots b_{n} 11}}^{2}\right)={\overline{11 b_{n} \ldots b_{2} b_{1}}}^{2}=2 \cdot 2^{n}+{\overline{1 b_{n} \ldots b_{2} b_{1}}}^{2} \\
& =3 \cdot{\overline{1 b_{n} \ldots b_{2} b_{1}}}^{2}-2 \cdot{\overline{b_{n} \ldots b_{2} b_{1}}}^{2}=2 g(2 n+1)-g(n)
\end{aligned}
$$

The number of fixed points is then easily calculated: half (in the broad sense) of the number determines the remainder, and the last digit must be 1. Using that $2011=11111011011^{2}$, this gives a total of $1+1+2+2+4+4+8+8+16+16+2 \cdot(14)=90$ fixed points.