[cross posted from math.se due to lack of answers]


I'm attempting to compute the group of continuous (or more generally holomorphic) $\phi: \mathbb{C} \rightarrow \mathbb{C}$ such that $f(\phi(x)) = f(x)$. 


In the linear setting, $f = ax + b$ we don't see anything interesting, the group $\phi$ is trivial.


In the quadratic setting, we note that $f = x^2$ has symmetry group $\lbrace x, -x \rbrace$, has $x^2 = (-x)^2$ and to go a step further, we can complete the square on any $ax^2 + bx + c$, to derive:


$$ ax^2 + bx + c = a \left(x^2 + \frac{b}{a} x + \frac{c}{a}\right) = a\left( x + \frac{b}{2a} \right)^2 + c - \frac{b^2}{4a} $$


And thus generally it has symmetry group $\lbrace x, -x-\frac{b}{2a} \rbrace$, 


Now the cubic case becomes much trickier. we can extrapolate from $f = x^3$ to the general $e\_0(x+e\_1)^3 + e\_2$ but this doesn't cover all possible cubic polynomials.


Any idea how to dig deeper here and compute the symmetries of all cubic polynomials?




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Some additional observations, in the quadratic case: the solutions of $ax^2 + bx + c = t$ must necessarily be permuted by such a symmetry, i.e. i was looking for a function that sends $-\frac{b}{2a} + \frac{\sqrt{b^2 - 4a(c-t)}}{2a} \rightarrow -\frac{b}{2a} - \frac{\sqrt{b^2 - 4a(c-t)}}{2a}$ 


In the cubic case, I can do a similar thing, writing down the form of the roots, then deducing ANY function that sends one form to the other, but such functions seem like they will be very unwieldy to write. And i'm not sure how to guess a form of such a function.



Computing the general symmetry group of cubic polynomials
For any polynomial $f$ of degree $n$, 
$$f(\phi(x)) - f(x) = (\phi(x) - x) g(\phi(x),x)$$ where $g$ is a bivariate polynomial of degree $n-1$, so the symmetries will be $\phi(x) = x$ and the roots of $g(\cdot, x)$. In general, the latter will not be entire functions. Thus if $f(x) = x^3 + a x$, we get
$$ \phi(x) = \frac{-x\pm\sqrt {-3\,{x}^{2}-4\,a}}{2} $$


