Is it true that any two of the graphs of cubic functions are similar? It is evident that any two lines are congruent. In addition, any two parabolas are similar. I wonder whether every graph defined by a cubic function $y=ax^3+bx^2+cx+d$ can be rewritten as $Y=X^3$ via a similarity transformation.
It can be easily demonstrated that a graph defined by a cubic function can be rewritten as $y=x^3+Ax$ via a similarity transformation. Therefore, the question boils down to whether the graph of $y=x^3+Ax$ is similar to that of $y=x^3$ for every real number $A$.
It seems counterintuitive to think the graph of $y=x^3-3x$ is similar to that of $y=x^3$ since the former has two &quot;prominent&quot; points. Nonetheless, I cannot prove or disprove the proposition that any two of the graphs of cubic functions are similar. Any helpful advice is cordially appreciated.
A short disclaimer/warning: I haven't studied these types of problems formally, so there may be a standard, simpler way to deal with this.  This is just the method that occurred to me.
Also, in the general proof, I abuse the symbol $y'$ a fair bit to mean the derivative on different curves, even in the same equation.  Hopefully the context gives enough clues to what I mean, but if anyone knows a standard notation for this application, I'd appreciate comments to that effect.

For the moment, let's consider $y=x^3$ and $y=x^3-3x$ as you have done.  Note that the range of the derivative of each function is of the form $[c_0,\infty)$ for different values of $c_0$ ($0$ and $-3$ respectively).  This difference is what we want to capture, but we need to define a quantity that both makes sense and is independent of similarity.
To do this, first we make the interval bounded by applying $\arctan$ so the range of $\arctan(y')$ is of the form $[c_1,\pi/2)$ for different values of $c_1=\arctan(c_0)$.  Then take [Lebesgue] measures, giving $$\lambda(\arctan(y'(\mathbb{R}))) = \frac{\pi}{2} - c_1.$$  This gives us the quantity we want.

More generally, we can consider any smooth curve $C \subset \mathbb{R}^2$ and let $[C]$ denote the equivalence class of all curves which are similar to $C$.  Then I claim that the function $$[C] \mapsto \lambda(\arctan(y'(C)))$$ is well-defined.  To see this, it's enough to see how it behaves under translation, uniform scaling, reflection, and rotation.

Translation and uniform scaling: Note that $y'(C)$ is unchanged under both, so obviously the above value is unchanged.
The particular reflection $R_y(x,y) = (x,-y)$ satisfies
$$\begin{align*}\lambda(\arctan(y'(R_yC))) &amp;= \lambda(\arctan(-y'(C))) \\ &amp;= \lambda(-\arctan(y'(C))) \\ &amp;= \lambda(\arctan(y'(C)))\end{align*}$$
and any other reflection can be written as $T\rho R_y \rho^{-1}T^{-1}$ where $T$ is a translation and $\rho$ is a rotation
For $0 \leq \theta &lt; \pi$, any counterclockwise rotation $\rho_{\theta}$ by $\theta$ about some point satisfies, for $(x,y) \in C$, $$\arctan(y'(\rho_\theta(x,y))) = \begin{cases}\arctan(y'(x,y)) + \theta &amp; \text{ if } \arctan(y'(x,y)) + \theta \leq \frac{\pi}{2} \\ \arctan(y'(x,y)) + \theta - \pi &amp; \text{ otherwise}\end{cases}$$ Using this, we may show directly that $\lambda(\arctan(y'(\rho_\theta C))) = \lambda(\arctan(y'(C))).$
For $\pi \leq \theta &lt; 2\pi$, just note $\rho_\theta =
   (\rho_{\theta/2})^2$ and $0 \leq \theta/2 &lt;\pi$.


Finally, applying the above to $C_A : y=x^3 + Ax,$ we find that $$\lambda(\arctan(y'(C_A))) = \frac{\pi}2 - \arctan(A)$$ so none of these curves are similar to one another.
There are four operations we can perform to convert two similar 2D graphs: translation, reflection, rotation, and uniform scaling
It is obvious that the first three of these cannot change the existence of local extreme points.
If we uniformly scale an equation by a factor of $\lambda$, we essentially replace $x$ with $\lambda x$ and $y$ with $\lambda y$.  If we take a cubic with no local extreme points, i.e. one of the form $y-A=C(x-B)^3$ and rescale it, we get $\lambda y-A= C (\lambda x-B)^3$.  If we redefine $A'=\frac{A}{\lambda}, B'=\frac{B}{\lambda}&lt;, C'=C\lambda^2$, this is equivalent to $y-A'=C'(x-B')^3$, which is also a cubic with no local extreme points