Take, for example, $f(x)=\frac{x-3}{x+1}$. One can verify that $f\circ f\circ f$ is the identity, so $f$ has order 3 in the group of Möbius transformations. Constructing such functions can be done easily.
Are there Möbius transformations of aribtrarily greater orders? If so, how can one construct them?
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Let $\zeta$ be a primitive $n^{\text{th}}$ root of unity, and consider the Möbius transformation $f(z) = \zeta z$. As
$$(\underbrace{f\circ f\circ\dots\circ f}_{k\ \text{times}})(z) = \zeta^kz,$$
the order of $f$ is $n$. Therefore, the group of Möbius transformations has an element of any finite order. In addition, $g(z) = z + 1$ provides an example of an element of infinite order.
The composition of Möbius transforms is naturally associated with their matrix of coefficients:
$$x \rightarrow f(x)=\dfrac{ax+b}{cx+d} \ \ \ \leftrightarrow \ \ \ \begin{bmatrix} a & b\\ c & d \end{bmatrix}$$
This correspondence is in particular a group isomorphism between the group of (invertible) homographic transforms of the real projective line and $PGL(2,\mathbb{R})$.
(composition $\circ$ mapped to matrix product $\times$).
Thus, your question boils down to the following: for a given $n$, does it exist a $2 \times 2$ matrix $A$ such that $A^n=I_2$ ?
The answer is yes for real coefficients. It suffices to take the rotation matrix :
$$\begin{bmatrix} \cos(a) & -\sin(a) \\ \sin(a) & \cos(a) \end{bmatrix} \ \ \ a=\dfrac{2\pi}{n}$$
Edit: If you are looking for integer coefficients, the answer is no. In fact, with integer coefficients, only homographies of order 2,3,4 and 6 can exist. (I rectify here an error that has been signaled and I add information). See for that the very nice paper (http://dresden.academic.wlu.edu/files/2017/08/nine.pdf) (in particular its lemma 1).