Let us denote by $F$ the set of real-valued functions of a real variable such that $$F=\left\{\frac{1}{1-x},\frac{x-1}{x},x\right\}.$$ We shall prove that this set forms an abelian group under the compositon of functions. We first notice that the set has closure and is commutative. Indeed, \begin{alignat*}{2} \left(x\right)\circ\left(\frac{1}{1-x}\right) &= \left(\frac{1}{1-x}\right)\in F &&= \left(\frac{1}{1-x}\right)\circ \left(x\right)\\ \left(x\right)\circ\left(\frac{x-1}{x}\right) &= \left(\frac{x-1}{x}\right)\in F &&= \left(\frac{x-1}{x}\right)\circ \left(x\right)\tag{1}\\ \left(\frac{1}{1-x}\right)\circ\left(\frac{x-1}{x}\right) &= \ \quad\left(x\right)\in F &&= \left(\frac{x-1}{x}\right)\circ \left(\frac{x-1}{x}\right). \end{alignat*} Now, we clain that $x$ is the identity element of the set. Indeed, \begin{alignat*}{2} \left(x\right)\circ\left(\frac{1}{1-x}\right) &= \left(\frac{1}{1-x}\right)\circ \left(x\right) &&= \frac{1}{1-x}\\ \left(x\right)\circ\left(\frac{x-1}{x}\right) &= \left(\frac{x-1}{x}\right)\circ \left(x\right) &&= \frac{x-1}{x}\tag{2} \end{alignat*} $$ \left(x\right)\circ\left(x\right) = x.$$
Moreover, we see from (1) and (2) that every element is invertible, hence the set forms an abelian group under the composition operator. We would now like to show that $F$ is a cyclic group:
Let us define the $n^{\text{th}}$ power $f^{n}$ of an element of $f\text{ of } F$ as the following
$$f^n =\begin{cases} e & : n = 0 \\ f^{n-1} \circ f & : n > 0 \\ \left({f^{-n}}\right)^{-1} & : n < 0 \end{cases}$$
where $e$ is the identity element of $F.$ We have to prove that every element in $F$ can be expressed as the power of a single element from $F$. We claim this generator is $\frac{1}{1-x}$. Indeed, $$\frac{1}{1-x}^{0}=x,\qquad \frac{1}{1-x}^{1}=\frac{1}{1-x},\qquad \frac{1}{1-x}^{2}=\frac{x-1}{x},\\$$ hence $F$ is the cyclic group of order 2 generated by $\frac{1}{1-x}$. Namely,
$$ F=\left\{\frac{1}{1-x}^{0},\frac{1}{1-x}^{1},\frac{1}{1-x}^{2} \right\}=\left<\frac{1}{1-x}\right>.\\$$ All cyclic groups of equal order being isomorphic to each other, consider the bijection $\phi\colon\mathbb{Z}/3\mathbb{Z}\to{F}\colon\phi(2^{n})=\frac{1}{1-x}^{n}$. We would like to show this is a homomorphism from $\mathbb{Z}/3\mathbb{Z}$ to $F$.
Note that $\phi(2^{n})=\frac{1}{1-x}^{n}$ holds for all $n$, for let $n$ be an integer such that $n = q k + r$ for $0 \le r < k$. Now
$$ 2^n = 2^r,\qquad \text{and}\qquad \frac{1}{1-x}^r = \frac{1}{1-x}^n,\\ $$ thus $$ \phi (2^{n}) = \phi (2^{r}) = \frac{1}{1-x}^r = \frac{1}{1-x}^n. $$ Now let $x,y$ be in $\mathbb{Z}/3\mathbb{Z}$. Since $\mathbb{Z}/3\mathbb{Z}=<2>,$ it follows that there exist integers $s,t$ such that $$x=2^{s} ,\qquad \text{and}\qquad y=2^{r}, $$ thus, \begin{align*} \phi(xy)&=\phi(2^{s}2^{t})\\ \qquad &=\phi(2^{s+t})\\ \qquad &=\frac{1}{1-x}^{s+t}\\ \qquad &=\frac{1}{1-x}^{s}\frac{1}{1-x}^{t}\\ \qquad &=\phi(2^{s})\phi(2^{t})\\ \qquad &=\phi(x)\phi(y) \end{align*} so $\phi$ is a homomorphism. As $\phi$ is a bijection, then it is also an isomorphism from $\mathbb{Z}/3\mathbb{Z}$ to $F$, hence $\mathbb{Z}/3\mathbb{Z} \cong F$, as desired. Lastly, let $h\colon \mathbb{N}\to \mathbb{Z}/3\mathbb{Z}$ be a function mapping a positive integer into its equivalence class modulo 3. Then
\begin{alignat*}{4} &\qquad \qquad \qquad \qquad \quad f_{2018}(x)&&=(\phi\circ h)(x)\\ & &&=\phi(h(2018))(x)\\ & &&=\phi(2)(x)\\ & &&=\phi(2)(x)=\frac{1}{1-x}^2\\ & &&=x\\ &\qquad \qquad \qquad \quad \implies f_{2018}(2018) &&=2018 \end{alignat*}
Another method:
Check that : $$f_0(x)=\frac {1}{1-x}$$
$$f_1(x)=1-\frac 1x$$
$$f_2(x)= x$$
And now hereafter it would be quite obvious that $$f_n(x) = \begin{cases} \frac{1}{1-x}, & \text{if $n=3k$} \\ 1-\frac 1x, & \text{if $n=3k+1$} \\ x, & \text{if $n=3k+2$} \end{cases}$$