Finding Symmetry Group $S_3$ in a function I was considering functions $f: \Bbb{C} \rightarrow \Bbb{C}$ and I defined the following instrument (I call it the Symmetry Group of a function)
$$ \text{Sym}(f) = \left< m(x)|f(m(x))=f(x) \right> $$
An intuitive example is to consider $\text{Sym}(e^x)$ and observe that
$$m(x) = x + 2i \pi $$ 
has the property that 
$$ e^{m(x)} = e^{x+2i\pi}=e^x e^{2i \pi} = e^x $$ 
And the group generated by $m(x)$ under composition is the set of functions 
$$ x + 2i\pi k, k \in \Bbb{Z}$$
Which is isomorphic to $\Bbb{Z}$ under function composition. So one can then say that $$\text{Sym}(e^x) \cong \Bbb{Z}$$
What I was curious about was if there are any elementary functions such that 
$$ \text{Sym}(g(x)) \cong S_3$$ 
In attempt to build one I considered 
$$ g(x) = x^{\frac{-1 + i \sqrt{3}}{2}} +  x^{\left( {\frac{-1 + i \sqrt{3}}{2}}\right)^2} + x + \frac{1}{x} +x^{-\frac{-1 + i \sqrt{3}}{2}}+  x^{-\left({\frac{-1 + i \sqrt{3}}{2}}\right)^2} $$
G has as a generator for its symmetries the functions $L_1 = \frac{1}{x}$ and $L_2 =  x^{\frac{-1 + i \sqrt{3}}{2}}$ 
Which can be observed as 
$$G(L_1) = G(L_2) = G(x)$$
But the problem is that $L_1(L_2) = L_2(L_1)$ so clearly this isn't a generating set for $S_3$. It's not obvious at this point, how to go about making a function that has $S_3$ as its underlying symmetry group

Some Examples:
$\Bbb{Z}_2$ can be realized as $\text{Sym}\left(x + \frac{1}{x}\right)$ as this function is invariant under the substitutions $x \rightarrow x$ and $x\rightarrow \frac{1}{x}$
The proof arises from the following: Suppose we wish to find all transformations $T$ $x$ such that 
$$ x + \frac{1}{x} = T(x) + \frac{1}{T(x)}$$
We can the derive that 
$$ T(x)^2 - \left(x+ \frac{1}{x}\right)T(x) + 1 = 0$$
Which yields that 
$$ T(x) = \frac{x + \frac{1}{x} \pm \sqrt{x^2+2+\frac{2}{x^2}-4}}{2}$$
simplifying to
$$ T(x)  = \frac{x + \frac{1}{x} \pm (x-\frac{1}{x})}{2}$$
and that gives $T(x) = x, T(x) = \frac{1}{x}$ observe the these transformations form a group of order $2$ so they must be isomorphic to $\Bbb{Z}_2$
And in general I conjecture that:
$\Bbb{Z}_n$ can be realized as $$\text{Sym} \left( x + x^{\sqrt[n]{1}_1} + x^{\sqrt[n]{1}_2} + ... x^{\sqrt[n]{1}_{n-1}}\right)$$

There's a little ambiguity as to the type of functions you are considering. For example, you seem ok with allowing the function to have some singularities at zero as your $z+1/z$ example indicates. The group $S_3$ acts naturally on $\mathbf{C} \cup \{\infty\}$ via the following rational functions:
$$\Sigma = \left\{z, \ 1/z, \  1-z,\frac{1}{1-z}, \ 1 - \frac{1}{z}, \ \frac{z}{z-1}\right\}$$
In particular, if $h(z)$ is any function $h: \mathbf{C} \cup \{\infty\} \rightarrow \mathbf{C} \cup \{\infty\}$ then 
$$f(z) = h(z) + h(1/z) + h(1-z) + h\left(\frac{1}{1-z}\right) + h\left(1-\frac{1}{z}\right) + h\left(\frac{z}{z-1}\right)$$
will be invariant under $\Sigma$. It could be the case that $f(z)$ is invariant under more symmetries, of course. For example, if $h(z) = z$, then $f(z) = 3$.
On the other hand, if $h(z) = z^2 + c$ for any constant $c$, then $f(z)$ is a non-trivial rational function. Moreover, one finds that (in this case)
$$f(x) - f(y) = \frac{2(x-y)(x+y-1)(xy - 1)(1-x+xy)(1-y+xy)(-x-y+xy)}{(x-1)^2 x^2 (y-1)^2 y^2}.$$
Assuming that $y \in \mathrm{Sym}(f)$, the numerator is zero, and so (under very weak continuity hypotheses) one of the six factors in the numerator are zero, leading to $y \in \Sigma$.  So it seems that $f(z)$ is a suitable function in your case. A particularly nice choice of constant $c$ is $c = -7/4$, in which case $f(2) = 0$, and so
$$f(z) = f(z) - f(2) = \frac{(z-2)^2 (z+1)^2 (2z - 1)^2}{2 z^2 (z-1)^2}.$$
In this case, the square-root of this function is invariant under the even elements of $\Sigma = S_3$ and sent to its negative under the odd elements.
A slightly more general nice family (but no longer a square) is given (for a parameter $t$) by
$$f(x) = \frac{2(x-t)(x+t-1)(xt - 1)(1-x+xt)(1-t+xt)(-x-t+xt)}{(x-1)^2 x^2 (t-1)^2 t^2}.$$

I might as well add a complete list of such examples coming from polynomials. Suppose that $f(x)$ is a polynomial, and $y \in \mathrm{Sym}(f)$. Then we must have $f(x) - f(y) = 0$. But $f(x) - f(y)$ is a rational function in $y$, and so has a finite number of algebraic solutions. If we insist that our functions are entire functions on $\mathbf{C} \cup \{\infty\}$, then this forces $y$ to be a rational function (other algebraic functions will not be single valued), and (by degree considerations) a function of the form:
$$y = \frac{a x + b}{c x + d}.$$
The choice of constants is only well defined up to scaling. This gives an injective map:
$$\mathrm{Sym}(f) \rightarrow \mathrm{PGL}_2(\mathbf{C}).$$
The finite subgroups of the right hand side are well known, and so, in particular, we deduce:
Claim: Let $f$ be a rational function. Then $\mathrm{Sym}(f)$ is either cyclic, dihedral, or one of the exceptional groups $A_4$, $S_4$, and $A_5$.
Cyclic examples are easy to construct. Let $f(x) = x^n$, and then $\mathrm{Sym}(f)$ consists of $y = \zeta x$ for an $n$th root of unity $\zeta$. This corresponds to the map:
$$a \in \mathbf{Z}/n \mathbf{Z} \mapsto \left( \begin{matrix} \zeta^a & 0 
\\ 0 & 1 \end{matrix} \right) \in \mathrm{PGL}_2(\mathbf{C}).$$
Naturally one can also take $f(x) = h(x^n)$ for a generic rational function $h(x)$.
Note that other examples (such as $f(x) = x + x^{-1}$) can be obtained from
these examples by suitable change of variables, namely, because
$$\left( \begin{matrix} 1 & - 1 \\ 1 & 1 \end{matrix} \right) 
\left( \begin{matrix}  0 & 1 \\ 1 & 0  \end{matrix} \right) 
\left( \begin{matrix} 1 & - 1 \\ 1 & 1 \end{matrix} \right)^{-1} = 
\left( \begin{matrix} -1 & 0  \\ 0 & 1 \end{matrix} \right),$$
and we find that
$$h(x) = x + \frac{1}{x}, \qquad h\left(\frac{x-1}{x+1}\right) = g(x^2), \qquad
g(x) = 2 \cdot \frac{x+1}{x-1}.$$
Note that the dihedral representation of $D_{2n}$ inside $\mathrm{PGL}_2(\mathbf{C})$ is
given by the image of $\mathbf{Z}/n \mathbf{Z}$ together with the matrix
$$\left( \begin{matrix}  0 & 1 \\ 1 & 0  \end{matrix} \right) ,$$
Hence we can write down the examples
$$f(x) = h\left(x^n + \frac{1}{x^n}\right),$$
for a generic function $h$ (taking $h(x) = x$ will do). Here $\mathrm{Sym}(f)$ is generated by $x \mapsto \zeta x$ and $x \mapsto 1/x$.
One can construct the other examples in a similar manner. For fun, I computed an example with $\mathrm{Sym}(f) = A_4$. The group $A_4$ has (several) projective representations
$$A_4 \rightarrow \mathrm{PGL}_2(\mathbf{C})$$
realized by $2$-dimensional representations of the Schur cover $\mathrm{SL}_2(\mathbf{F}_3)$.
One such example maps the non-trivial elements of the 
 Klein $4$-subgroup $K$ to
$$K \setminus \{e\} = \left\{\left( \begin{matrix}  0 & 1 \\ 1 & 0  \end{matrix} \right), \left( \begin{matrix} -1 & 0  \\ 0 & 1 \end{matrix} \right), \left( \begin{matrix} 0 & -1  \\ -1 & 0 \end{matrix} \right)\right\},$$
and this group is normalized by the order three (in $\mathrm{PGL}_2$) element
$$\left( \begin{matrix}i  & -i  \\ 1 & 1 \end{matrix} \right)$$
Writing down the corresponding $12$ elements of $A_4$ and letting
$$f(z) = \sum_{A_4 \subset \mathrm{PGL}_2(\mathbf{C})} h(\gamma z),$$
doing a calculation
as above with $h(z) = z^2 + c$, one finds that
$$\begin{aligned}
f(x) - f(y) = & \ 2(x - y)(x + y)(-1 + xy)(1 + xy)(-i - ix - y + xy)(i + ix - y + xy)
  (-i - x - iy + xy)\\
 \times & \ 
\frac{(i + x - iy + xy)(i - x + iy + xy)(-i + x + iy + xy)
  (i - ix + y + xy)(-i + ix + y + xy)}{(-1 + x)^2x^2(-i + x)^2(i + x)^2
  (1 + x)^2(-1 + y)^2y^2(-i + y)^2(i + y)^2(1 + y)^2} \end{aligned}
$$
Since translating $f(x)$ preserves the symmetry group, one can (for example) choose $f(x)$ to vanish at $x = y$ for any fixed $y$, and then $f(x)=f(x) - f(y)$ as above. For example, if $y = 2$, then
$$450 f(x) = \frac{(-2 + x) (2 + x) (-1 + 2 x) (1 + 2 x) (9 + x^2) (5 - 6 x +    5 x^2) (5 + 6 x + 5 x^2) (1 + 9 x^2)}{(-1 + x)^2 x^2 (1 + 
   x)^2 (1 + x^2)^2}.$$
