Help with operator $f(x^q)=\frac{1}{q+1}x^q$. This question is somewhat related to this.
I am looking for an operator $f:\mathbb{R}[x]\to\mathbb{R}[x]$, that is, $f$ is an operator that maps polynomials in one variable to polynomials in one variable.
The requirement of $f$ is that it must map the monomial $x^q$ to the monomial $\frac{1}{q+1}x^q$, that is $f(x^q)=\frac{1}{q+1}x^q$. This operator should be independent of $q$.
I first thought on using integration, however I do not want this (see below). 
My proposal is the following: 
Let $P\in\mathbb R[x]$. We define $\phi:\mathbb R[x]\to\mathbb R[x]$ by $\phi(P)=\frac{d}{dx}(xP)$ (I think I can freely use $x$). 
Next we define $f:\mathbb R[x]\to\mathbb R[x]$ by $f(P)=\frac{P^2}{\phi(P)}$. Note that in the particular case $P=x^q$ we do satisfy our requirement $$f(x^q)=\frac{x^{2q}}{(q+1)x^{q}}=\frac{1}{q+1}x^q.$$
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Now, let us extend the previous idea to operators $f:\mathbb R[x_1,\ldots,x_n]\to\mathbb R[x_1,\ldots,x_n]$ . 
Let $P\in \mathbb R[x_1,\ldots,x_n]$. Following the above stuff we define a family of maps $\phi_k:\mathbb R[x_1,\ldots,x_n]\to\mathbb R[x_1,\ldots,x_n]$ by $\phi_k(P)=\frac{d}{dx_k}(x_kP)$ (again I think I can freely use $x_k$, also $\phi_k(\cdot)=\phi(\cdot,k) $ with $k\in\mathbb N\backslash \left\{ 0 \right\}$). 
Next, we define similarly $f_k(P)=\frac{P^2}{\phi_k(P)}$. 
In the particular case that $P=x^Q=x_1^{q_1}\cdots x_n^{q_n}$, we have
$$
f_k(x^Q)=\frac{(x_1^{2q_1}\cdots x_k^{2q_k}\cdots x_n^{2q_n}) }{ (q_k+1)(x_1^{q_1}\cdots x_k^{q_k}\cdots x_n^{q_n}) }=\frac{1}{q_k+1}x^Q.
$$
What do you think?, any criticism is appreciated. Specially: is it true that if I define an operator on $P$, I can use $x_k$?

If $x\in\mathbb{R}$ it is OK to define $f(x^q)=x\int x^q dx=\frac{1}{q+1}x^q+c$.
However if $x\in\mathbb{R}^n$ then we would have $f_k(x^q)=x_k\int x^q dx_k=\frac{1}{q+1}x^q+\phi(x_1,\cdots,x_{k-1},x_{k+1},\cdots,x_n)$, where $\phi$ is a function of all terms $x_j$ except $x_k$.  

EDIT: I'm quite sure that I misunderstood the OP's intention. However, the comments to this answer may still be of interest and so have made this a wiki answer.
No such function exists. We want  $f(x^q)=x^q/(q+1)$ for all $q$; consider in particular $q=6$. Then we require \begin{align}
f(x^{6})
&=f((x^2)^3))=f((x^3)^2))\\
&=\frac{x^{6}}{7}=\frac{(x^{2})^3}{3}=\frac{(x^{3})^2}{4}\end{align}
which cannot be true since $7\neq 4\neq 3$.
