I'm not sure how to explain this too well so I am going to include some examples to explain what I mean by "isolate".
For example if I wanted to isolate all of the terms in $f(x)$ with even exponents I could simply use this expression: $$\frac{f(x)+f(-x)}{2}$$
More generally if I wanted to isolate all of the terms of $f(x)$ with an exponent divisible by $n$ I could use the following expression:
$$\frac{\sum_{k=0}^{n-1} f(e^{i \cdot 2\pi k/n} x) }{n}$$
My question is how could you extend this so that you could isolate all the terms with exponents that are $m \pmod n$ for any $m<n$
How about $$\frac{\sum_{k=0}^{n-1} e^{-i \cdot 2\pi km/n} f\left(e^{i \cdot 2\pi k/n} x\right) }{n}$$
Rationale: $g(x)=x^{n-m}f(x)$ has the same coefficients as $f$, shifted to the right by $n-m$. So if you apply your formula to $g$, you'll "isolate" the coefficients that $0$ (mod) $n$ for that polynomial. $$\frac{\sum_{k=0}^{n-1} g\left(e^{i \cdot 2\pi k/n} x\right) }{n}=\frac{\sum_{k=0}^{n-1} e^{-i \cdot 2\pi km/n} x^{n-m}f\left(e^{i \cdot 2\pi k/n} x\right) }{n}$$
Those are the coefficients that are $m$ (mod) $n$ for $f$, except that to get the right degrees back, you need to divide the result by $x^{n-m}$.