Vector space endomorphisms in $\mathbb{R}[x]$ commuting with $E:f\mapsto f+f'$ I am wondering if every vector space endomorphism in $\mathbb{R}[x]$ that commutes with $E:f\rightarrow f+f'$ is invertible. (denoting $f'$ the derivative of $f$)
To begin with, $E$ is invertible because if $D: f\rightarrow f'$, then for every polynomial $f\in\mathbb{R}[x]$ there exists a integer $n$ such that $D^nf=0$, namely $n=degf+1$. Thus, the inverse of $E$ can be described as $I-E+E^2-E^3+\cdots\pm E^n$.
If an endomorphism $A$ commutes with $E$, then $A(f+f')=Af+(Af)'$. Hence, all endomorphisms of the form $f\rightarrow af, \ a\in\mathbb{R}$ commute with $E$ and they are all invertible. We can think about endomorphisms of other forms, for example, $f\rightarrow \int_0^1 f$ doesn't commute with $E$ because for $f(x)=x^2$ we get $\int_0^1 (x^2+2x)=\frac{4}{3}$ but $\int_0^1 x^2+\Big( \int_0^1 x^2\Big)'=\frac{1}{3}$. 
I'm trying to find a way to "organise" all possible forms of endomorphisms in $\mathbb{R}[x]$ in order to check if they commute with $E$. My first thoughts were that all endomorphisms end up to have a similar form with the one of $E$, meaning that they decrease or increase the degree of the monomials of $f$ respecting linearity and add a multiple of $f$, but I don't know how to prove this.
Any hints concering a counterexample or a proof would be very helpful.
 A: If you mean vector space endomorphism, then consider $\frac{\partial}{\partial x}$. This linear operator commutes with the given one $E=1+\frac{\partial}{\partial x}$ but is not invertible. In fact, $C_{\mathrm{End}(\mathbb{R}[x])}(1+\frac{\partial}{\partial x})=\mathbb{R}[[\frac{\partial}{\partial x}]]$, i.e. every endomorphism commuting with $1+\frac{\partial}{\partial x}$ is an infinite power series in $\frac{\partial}{\partial x}$ with constant coefficients.
To see this, first note every endomorphism of $\mathbb{R}[x]$ may be written as $C=\sum_{k\ge0}c_k(x)\left(\frac{\partial}{\partial x}\right)^k$ for some arbitrary polynomials $c_0(x),c_1(x),\cdots$. To see this, observe that we may apply $C$ to the basis powers $1,x,x^2,\cdots$ in order to create a linear system of equations which we may solve recursively for the polynomials $c_k(x)$ in terms of $F(x^k)$ and previous $c_0(x),\cdots,c_{k-1}(x)$.
Then applying the commutator $[a,b]=ab-ba$ yields
$$\left[1+\frac{\partial}{\partial x},C\right]=\sum_{k\ge0}\left[1+\frac{\partial}{\partial x},c_k(x)\left(\frac{\partial}{\partial x}\right)^k\right]=\sum_{k\ge0}c_k'(x)\left(\frac{\partial}{\partial x}\right)^k=0.$$
Thus, $c_k'(x)=0$ implies each $c_k$ is constant, Q.E.D.
Infinite power series are invertible are precisely when they have nonzero constant coefficient.

My original answer when I thought the question was about algebra homomorphisms:
Any unital $\mathbb{R}$-algebra homomorphism $\mathbb{R}[x]\to R$ is determined by where $x$ is sent.
Say $\phi$ sends $x$ to $g(x)$. Then we may say that $\phi(f(x))=f(g(x))$ for all $f(x)\in\mathbb{R}[x]$.
The commutativity condition reads $(1+\frac{\partial}{\partial x})\phi(f(x))=\phi((1+\frac{\partial}{\partial x})f(x))$. This is
$$f(g(x))+g'(x)f'(g(x))=f(g(x))+f'(g(x)).$$
Thus $g'(x)=1\implies g(x)=x+h$ and $\phi(f(x))=f(x+h)$ for some constant $h$.
So inverse endomorphism is then given by $\phi^{-1}(f(x))=f(x-h)$.
To connect this back with the above part of this answer, note the Taylor series 
$$f(x+h)=\exp\left(h\frac{\partial}{\partial x}\right)f(x)=\left[\sum_{k\ge0}\frac{h^k}{k!}\left(\frac{\partial}{\partial x}\right)^k\right]f(x).$$
