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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
$$f=p^{-1}circ gcirc p.$$
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 $fsim T_n$ 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 $mathbb{R}^ntomathbb{R}^n$ 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.

Edit: In the particular case where we are looking at the Chebyshev polynomials, by definition $T_n(cos x)=cos(nx)$. Hence with $c:[0,pi)to[-1,1),xmapstocos x$ and $m:[0,pi)to[0,pi)$, $xmapsto nx$ mod $pi$, we have $T_ncirc c=ccirc m$, i.e. $T_n=ccirc mcirc c^{-1}$. Therefore $T_nsim m$, which is a linear map. Now $fsim T_n$ iff it is similar to the linear map $xmapsto nx$.

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 $xmapsto x-k$, then we have $phi^{-1}circ fcircphi$ 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 $phi:xmapsto kx$ 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 $partial qleqpartial p-1$ ($partial$ 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 $fnsim g$ by an affine bijection $phi$:

1. 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;


2. 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 $fsim g$ with affine $phi$. I am however unable to prove this. If this is true, then it gives us a test for conclusively saying $fsim g$ in the general case.