When is One Polynomial "Similar" to Another Let $f$ and $g$ be functions. We will say they are similar (in somewhat of an extension to what it means for linear maps to be similar) if there exists a bijection $p$ such that
$$f=p^{-1}\circ g\circ p.$$
Let this relation be denoted as $\sim$. Under what conditions can polynomials be similar? Specifically, I am trying to find out for which polynomials $f$ can we say that $f\sim T_n$ for some $n$ where $T_n$ is the $n$-th Chebyshev polynomial.
Note that when $f$, $g$, and $p$ are all restricted to linear functions from $\mathbb{R}^n\to\mathbb{R}^n$ this aligns with the normal definition of similar matrices. Could a similar equivalence be drawn for polynomial functions? Possibly one could consider matrices over the field $\mathbb{R}[x]$, and apply a similar theory to get partial results for the above problem.
 A: Edit: In the particular case where we are looking at the Chebyshev polynomials, by definition $T_n(\cos x)=\cos(nx)$. Hence with $c:[0,\pi)\to[-1,1),x\mapsto\cos x$ and $m:[0,\pi)\to[0,\pi)$, $x\mapsto nx$ mod $\pi$, we have $T_n\circ c=c\circ m$, i.e. $T_n=c\circ m\circ c^{-1}$. Therefore $T_n\sim m$, which is a linear map. Now $f\sim T_n$ iff it is similar to the linear map $x\mapsto nx$.

In general: let's exclusively discuss the case where $\phi$ is affine. If $f(x)=\sum a_ix^i$, then taking $\phi$ to be a translation $x\mapsto x-k$, then we have $\phi^{-1}\circ f\circ\phi$ to be $f(x-k)+k=\sum a_i(x-k)^i+k$. Hence, starting from any polynomial $f$, shifting its graph along the diagonal will always result in a similar function. Furthermore, the choice $\phi:x\mapsto kx$ will result in the similar function $f(kx)/k$, so we conclude that "shrinking" the graph of $f$ in a ratio-preserving manner will also result in the graph of a similar function.
Any polynomial of odd degree crosses the diagonal $y=x$ at least once, so bring this point to the origin (obtaining a similar function). By suitable scaling, we conclude that any odd degree polynomial $p(x)$ is similar to $xq(x)$, where $\partial q\leq\partial p-1$ ($\partial$ denotes the degree). Furthermore, it's not hard to show that similar functions must cross the diagonal the same number of times. This gives us several methods to prove that $f\nsim g$ by an affine bijection $\phi$:


*

*If the number of roots of $f(x)-x$ is not the same as the number of roots of $g(x)-x$, then they are not similar;

*For all the roots $x_k$ of $g(x)-x$, look at the similar monic polynomials $g'_k(x)=xh_k(x)$ defined by "bringing the root to the origin" and scaling such that the leading coefficient is $1$. Do the same for $f$. If the resulting monic polynomials are not all the same, then they are not similar.


I believe these are exhaustive; i.e. if $f$ passes both of these tests, then $f\sim g$ with affine $\phi$. I am however unable to prove this. If this is true, then it gives us a test for conclusively saying $f\sim g$ in the general case. 
