Let $\varphi: \mathbb{R}[x] \to \mathbb{R}[x],f(x) \mapsto f(2x+1)$. To determine the eigenvalues, I have:
Let $b_i = x^{i}$ be the i. basis-vector of $\mathbb{R}[x]$. $\varphi(b_i) = (2x+1)^i = \sum_{k=0}^i \binom{i}{k} 2x^{i-k} = b_i \cdot \sum_{k=0}^i \binom{i}{k} 2x^{-k}$. How can I determine the eigenvalues? I know, that $f(2x+1)=\lambda \cdot f(x)$, but how can I get rid of $x$ in $\sum_{k=0}^i \binom{i}{k} 2x^{-k}$, as $f(2x+1) = \sum_{i=0}^n a_i (2x+1)^i = \sum_{i=0}^n a_i \sum_{k=0}^i \binom{i}{k} 2x^{i-k}$.
Say $f$ is a degree-$n$ eigenpolynomial of $\varphi$. Then $$ f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0\\ \varphi(f)(x) = a_n(2x+1)^n + a_{n-1}(2x+1)^{n-1} + \cdots + a_1(2x+1) + a_0 $$ By looking at the leading term of $\varphi(f)$, we see that the coefficient has changed from $a_n$ to $2^na_n$. So the eigenvalue corresponding to the degree-$n$ eigenpolynomial $f$ is $2^n$.
So the remaining question becomes: Is there an eigenpolynomial of each degree? I claim that $(x+1)^n$ is such an eigenpolynomial. Indeed, we have: $$ \varphi((x+1)^n) = (2x+1+1)^n = 2^n(x+1)^n $$
(I found this eigenpolynomial by solving explicitly for $n = 0, 1, 2$ and then confirming once for $n = 3$, before starting to expand the binomials to try to actually prove that $(x+1)^n$ worked in general. I was stuck for a minute staring at the sums, then I came to the realization above, which made the proof very simple.)