Consider the sets of degree $n$ polynomials, $$P_{n} = \big\{ a_n x^n + a_{n-1}x^{n-1} + \cdots a_1 x + a_0\ : a_{n} \neq 0 \big\},$$ and the collection of classical function transformations:
- horizontal dilation: $p(x) \mapsto p(x/k), \quad \text{where $k\neq 0$} $
- vertical dilation: $p(x) \mapsto k p(x), \quad \text{where $k \neq 0$}$
- horizontal translation: $p(x) \mapsto p(x-k)$
- vertical translation: $p(x) \mapsto p(x) + k$
It's clear that any composition of these transformations will map $P_n$ to itself. Moreover, it seems reasonable that these operations should allow us to define a set of congruence classes on the set $P_n$ defined as follows:
For $p(x)$ and $q(x)$ in $P_n$, we say that $p(x) \sim q(x)$ if and only if there is a set of function transformations which transform $p(x)$ into $q(x)$.
Obviously this relationship is reflexive, since $p(x) \sim p(x)$ by the identity transformation; and since each transformation has an inverse transformation, then the symmetric and transitive properties of $\sim$ should hold as well.
This brings me to my actual question:
What (if anything) can be said about the set of equivalence classes, $P_{n} \,/ \sim$?
Of note, since all parabolas are similar this means, interestingly enough, that $\left|P_{2}\, / \sim\right| = 1$, since the equivalence class $\left[x^2\right]$ contains all quadratics. I'm just not sure how to even approach something like $P_3 \,/ \sim$ or beyond, so any suggestions would be quite welcomed.
Only the horizontal transformations affects the position of the critical points $\{ x : p'(x) = 0 \}$; horizontal dilation dilates them, and horizontal translation translates them. This means that, for example, if $\deg p = 4$ and $p'$ has all real roots, then generically $p$ has three real critical points $c_1 < c_2 < c_3$ and so all four transformations leave the ratio
$$\frac{c_3 - c_1}{c_2 - c_1}$$
invariant. More generally, if $p$ has $k$ distinct real critical points $c_1 < c_2 < \dots < c_k$ then all four transformations leave invariant the point in projective space with coordinates $(c_k - c_1 : c_{k-1} - c_1 : \dots c_2 - c_1)$ (or, if you prefer, they leave invariant all ratios of the form $\frac{c_i - c_j}{c_k - c_{\ell}}$).
Conversely, if $p$ and $q$ are two polynomials of the same degree and their complex critical points (including multiplicity) have the property that it's possible to dilate and translate one set to become the other, then $p \sim q$. But unlike the real critical points, the complex critical points don't come in any preferred order, so it's not immediately clear to me how to translate this criterion into the equality of some more concrete set of invariants. In any case the passage to the (multi)set of critical points has the effect of ignoring the vertical transformations so one is left only to contemplate the effect of the horizontal ones. You can use this to write down a normal form by translating so that the complex critical points add up to $0$, then dilating so that e.g. the largest critical point in absolute value has absolute value $1$ or something like that.
This line of reasoning correctly suggests that the quadratic case is somewhat degenerate, since when $\deg p = 2$ there is only one critical point (the vertex of the parabola) and any two such critical points can be translated to each other.
The cubic case is also somewhat degenerate but less so. When $\deg p = 3$ there are two critical points so if they are distinct and real any two pairs of such critical points can be translated and dilated into each other. This suggests that the cubic case should split into a finite set of cases based on whether the roots of $p'(x)$ are distinct or real, and in fact you can show that any cubic is equivalent to exactly one of the cubics $x^3 + x, x^3, x^3 - x$, corresponding to the critical points being complex, there being a single critical point of multiplicity $2$, or the critical points being distinct and real. (Vertically dilate and translate to set the $x^3$ coefficient to $1$ and the constant coefficient to $0$, horizontally translate to set the $x^2$ coefficient to $0$, then horizontally dilate to normalize the $x$ coefficient.)
You can guess ahead of time that the quadratic and cubic cases are degenerate compared to the others because you have a $4$-parameter family of transformations (formally, a $4$-dimensional Lie group) whereas the polynomials of degree $n$ form an $n+1$-parameter family, so heuristically the transformations can successfully cut down the continuous variations of the polynomials iff $n+1 \le 4$ and this is what we observe.