Consider the set of three quadratic polynomials:
$a_1(x) = x^2 - 4x - 13$
$a_2(x) = x^2 + 3x - 2$
$a_3(x) = x^2 - x - 4$
Obviously, any two quadratics can be considered "similar" in that we can find a linear transformation $\phi(x)=ax+b$ such that $c\cdot a_1(x) = a_2(\phi(x))$, where $c$ is a scalar. We could prove this by expanding $a_2(\phi(x))$, but it's more intuitive to just consider that we can map the roots of $a_1$ onto the roots of $a_2$ by scaling and shifting appropriately.
I want to find a formula for $\phi$ and see when it is still valid even for higher degree polynomials.
Specifically, if we let $f(x) = x^n + b_{n-1}x^{n-1} + \cdots + b_0$ be a monic degree $n$ polynomial, we can define $\mu_f = \frac{-b_{n-1}}{n}$ to be the mean of the roots of $f$ by Viete's formulas. That is, $f(x)$ also equals $(x-r_1)(x-r_2)\cdots(x-r_n)$ when expanded, and we can match coefficients to see that $\Sigma r = -b_{n-1}$. We can also define $\sigma_f^2 = \frac{(n-1)b_{n-1}^2 - 2n b_{n-2}}{n^2}$ to be the variance of the roots. ($\sigma_f^2$ can be defined as the mean of the squares of the roots minus the square of the mean. We know the mean and hence the square of the mean, and finding the mean of the squares can be done by taking $-b_{n-1}$ which is the sum of the roots, squaring it and subtracting $2b_{n-2}$ which is the sum of the product of all pairs of roots, and then dividing $b_{n-1}^2-2b_{n-2}$ by $n$.)
We then can construct a candidate $\phi$ from $f$ to $g$ where $f$ and $g$ are both monic polynomials of degree $n$ in a couple steps:
First, shift so the mean is at $0$, ie $x - \mu_f$.
Next, scale so the variance is correct, ie $\pm\sqrt{\frac{\sigma_g^2}{\sigma_f^2}}(x - \mu_f)$.
Finally, shift so the mean is correct, ie $\mu_g\pm\sqrt{\frac{\sigma_g^2}{\sigma_f^2}}(x - \mu_f)$.
It's obviously necessary for a linear transformation such that $c\cdot f(x) = g(\phi(x))$ to have this form, since such a linear transformation must get the mean and variance of the roots correct, but it is very possible for the transformation constructed in this way to not actually give us the $c\cdot f(x) = g(\phi(x))$ similarity we want. For quadratics, the roots are defined by their mean and variance, but for degree $n$ polynomials in general there are still $n-2$ degrees of freedom in the roots even with mean and variance fixed, yet linear transformations only give us the power to match these first two degrees of freedom.
In the example case with $a_1$, $a_2$, and $a_3$, we can check that this construction works by constructing a linear transformation from $a_1$ to $a_2$ and one from $a_2$ to $a_3$. The remaining four transformations from $a_i$ to $a_j$ could then be obtained by inverting or composing those two.
First, we compute $\mu_{a_1} = 2$, $\mu_{a_2} = \frac{-3}{2}$, $\mu_{a_3} = \frac{1}{2}$, $\sigma_{a_1}^2 = 17$, $\sigma_{a_2}^2 = \frac{17}{4}$, and $\sigma_{a_3}^2 = \frac{17}{4}$.
Now, to go from roots of $a_1$ to roots of $a_2$, we compute $\phi(x) = \frac{-3}{2} \pm \frac{1}{2}(x - 2)$, and we can compute $a_2(\phi(x))$ to check it makes $a_1$ and $a_2$ "similar":
$a_2(\frac{-3}{2} \pm \frac{1}{2}(x - 2)) = \frac{9}{4} \pm 2\frac{-3}{2}\frac{1}{2}(x - 2) + \frac{1}{4}(x - 2)^2 + \frac{-9}{2} \pm \frac{3}{2}(x - 2) - 2 = \frac{1}{4}x^2 - x - \frac{13}{4} = \frac{1}{4}a_1(x)$.
Next, to go from roots of $a_2$ to roots of $a_3$, we compute $\phi(x) = \frac{1}{2} \pm (x + \frac{3}{2})$ and check $a_3(\phi(x))$:
$a_3(\frac{1}{2} \pm (x + \frac{3}{2})) = x^2 + 3x - 2 = a_2(x)$.
Notice that in the quadratic case, $\mu_f$ and $\sigma_f^2$ can be readily recombined to give the quadratic formula, and indeed the discriminant is $2\sigma_f$.
Now consider the following four cubic polynomials:
$b_1(x) = x^3 - x^2 - 10x + 8$
$b_2(x) = x^3 + 4x^2 - 5x - 16$
$b_3(x) = x^3 - 8x^2 - 51x - 58$
$b_4(x) = x^3 + 8x^2 - 20x - 128$
These have discriminant 3844, 3844, 61504, and 246016 respectively. These are all multiples of $3844=4\cdot 31^2$, but an even more obvious fact is that they are not the same. While the first two can be expressed in terms of one another using $\phi(x) = -1 - x$ (which is its own inverse), the other two cannot and so are not "similar" in the sense defined above.
I have two questions about this similarity:
Is there a sufficient condition for when two polynomials will be similar (especially a necessary and sufficient condition)? I could develop such a condition based on all the moments of the roots and Viete's formulas, since the mean and variance are essentially the first two moments and so we can say two polynomials are similar iff the remaining moments "line up" under the transformation which yields the right values for the first two moments, but this is pretty unwieldy. I hope there is some condition involving the discriminant, but I doubt it.
Is there a more general way in which these last four polynomials ARE all similar? The context in which they arose makes me expect they should be related somehow but aside from the determinants all being multiples of 3844 I don't really see it.
These polynomials were obtained by taking one monic, integral, irreducible degree 15 polynomial with 15 real roots, splitting its roots into 3 sets of 5, and constructing the monic, real polynomial with each set of roots. Then these 4 polynomials are the polynomials whose ROOTS are the COEFFICIENTS of those 3 polynomials for the 3 sets of 5 roots.
The first example of 3 quadratic polynomials were obtained in a similar way, except starting with a monic, integral, irreducible degree 8 polynomial with 8 real roots and splitting its roots into 2 sets of 4.