Linear operator on polynomials and invariant sets of roots

Question

Let $T:\mathbb{C}_n[x] \to \mathbb{C}_n[x]$ be a linear map from the vector space of polynomials of degree $n$ to itself.

Let $S \subset \mathbb{C}$ be a set with at least $3$ points, such that for every polynomial $P$ of degree $n$ with all roots in $S$, then $T(P)$ is either a constant, or has also all roots in $S$.

If $S$ is a finite set, is it necessarily the case that $T[p(x)] = p(\phi(x))$ for some affine map $\phi: \mathbb{C} \to \mathbb{C}$, or a Mobius map as Robert indicate below?

Can we even conclude that $T$ must be invertible?

Answer 1

Try $T f(x) = f(-x)$, and $S = \{-1, 0, 1\}$ (or any finite subset of $\mathbb C$ that is invariant under multiplication by $-1$).

EDIT: Of course, the vector space is polynomials of degree $\le n$, not $=n$.

Try $Tf(x) = x^n f(1/x)$, with $S$ the union of $\{0\}$ and the $m$'th roots of unity.

Still more generally, let $g(z) = (a z + b)/(c z + d)$ be a Möbius transformation of finite order (i.e. such that the $m$-fold composition $g^m$ is the identity), take $T f(x) = (c x + d)^n f(g(x))$, and $S \cup \{\infty\}$ a set invariant under $g$.

And if you want something more general than that, I don't know.

Answer 2

For $l$ a linear form on this space take $T = j\circ l$ where $j$ is the inclusion of $\mathbb C$ into constant polynomials.

This is not invertible.