I challenged myself to find a general formula for an $n$-degree polynomial with $n-2$ inflection points, all on the $x$-axis. Here is what I came up with (explanation is at the end).
$$\text{Even }n:f_n(x)=1+\sum_{k=1}^{n/2}\left(x^{2k}\prod_{i=1}^k \frac{(2i-2)(2i-3)-n(n-1)}{2i(2i-1)}\right)$$
$$\text{Odd }n: g_n(x)=x+\sum_{k=1}^{(n-1)/2}\left(x^{2k+1}\prod_{i=1}^k \frac{(2i-1)(2i-2)-n(n-1)}{2i(2i+1)}\right)$$
For example, here is the graph of $y=f_8(x)=\frac{1}{5}(429x^8-924x^6+630x^4-140x^2+5)$.
Then I discovered that these polynomials have another interesting property.
On a whim, for even $n$, I drew the graph of $y=f_n(x)|f_n(x)|$, which is like squaring the function but preserves positive and negative. As $n\to\infty$, the turning points approach a circle.
For odd $n$, the turning points on the graph of $y=n^2 g_n(x)|g_n(x)|$ (note the $n^2$) approach a circle.
(Another nice feature is that the turning points appear to be uniformly spaced around the circle.)
My question is:
Prove that, as $n\to\infty$, the turning points on $y=f_n(x)|f_n(x)|$ and $y=n^2 g_n(x)|g_n(x)|$ approach a circle.
(I think we will see the same phenomenon with any $n$-degree polynomial whose $n-2$ inflection points are all on the $x$-axis, where $n$ is large. That is, if we take any such polynomial, and multiply it by its modulus, and apply a certain vertical stretch, then the turning points will be approximately on a circle, uniformly spaced.)
Here is how I derived $f_n(x)$. Assume $f_n(x)=\sum\limits_{k=0}^n a_k x^k$ with leftmost and rightmost roots at $x=\pm1$, and $a_0=1, a_1=0$. To ensure that the inflection points are all on the $x$-axis, let $f_n(x)=(x-1)(x+1)\frac{{f_n}''(x)}{n(n-1)}$. Equate coefficients, then the above expression for $f_n(x)$ follows. The same method can be used to derive $g_n(x)$, except we assume $a_0=0$ and $a_1=1$.
Possibly related: question about an $(n+1)$-degree polynomial that is tangent to a circle at $n$ points.
EDIT
My derivation of $f_n(x)$ and $g_n(x)$ is incomplete, because I have not shown that they each have $n-2$ inflection points. I do not know how to show this.
For even $n$, let $p_n(x)=(-1)^{n/2}(1-x^2)^{1/4}\cos{(n\arccos{x})}$.
($\cos{(n\arccos{x})}$ is the $n$th Chebyshev polynomial of the first kind.)
$\color{red}{\text{We will show that the turning points on $y=p_n(x)|p_n(x)|$ are close to a circle.}}$
Here is the graph of $y=p_n(x)$ for $n=16$, and $y=\pm(1-x^2)^{1/4}$.
$y=\cos{(n\arccos{x})}$ is tangent to $y=\pm1$ at $x=\cos{(\frac{k\pi}{n})}, 0<k<n$.
So $y=p_n(x)$ is tangent to $y=\pm(1-x^2)^{1/4}$ at $x=\cos{(\frac{k\pi}{n})}, 0<k<n$. For large $n$, the turning points on $y=p_n(x)$ are close to these tangent points, so the turning points on $y=p_n(x)$ are close to $y=\pm(1-x^2)^{1/4}$.
So for large $n$, the turning points on $y=p_n(x)|p_n(x)|$ are close to $y=\pm(1-x^2)^{1/2}$, which is a circle.
$\color{red}{\text{We will show that, for the OP's $f_n(x)$, we have $f_n(x)\approx p_n(x)$ for large $n$.}}$
A tedious but straightforward calculation shows that $\lim\limits_{n\to\infty}\dfrac{(x^2-1){p_n}''(x)}{n(n-1)p_n(x)}=1$.
$\therefore p_n(x)\approx \dfrac{(x^2-1){p_n}''(x)}{n(n-1)}$ for large $n$.
The OP showed that there is a unique polynomial with $a_0=1, a_1=0$ satisfying $f_n(x)=\dfrac{(x^2-1){f_n}''(x)}{n(n-1)}$.
$\therefore f_n(x)\approx p_n(x)$ for large $n$.
$\color{red}{\text{$\therefore$ For large $n$, the turning points on $y=f_n(x)|f_n(x)|$ are close to a circle.}}$
We can use a similiar argument for odd $n$. Let $q_n(x)=\frac{1}{n}(-1)^{(n-1)/2}(1-x^2)^{1/4}\cos{(n\arccos{x})}$ and show that the turning points on $y=n^2 q_n(x)|q_n(x)|$ are close to a circle. Then show that $g_n(x)\approx q_n(x)$ for large $n$.
EDIT
I found a flaw in my answer. To reach the step where I wrote "$\therefore f_n(x)\approx p_n(x)$", I assumed that $p_n(x)$ is a polynomial of degree $n$, but this assumption is wrong. I do not know how to fix this flaw.