Can anyone find a polynomial $f ∈ ℚ\left(X+\frac{1}{1-X} + \frac{X-1}{X}\right) ⊆ ℚ(X)$ that is fixed under the automorphism $(X ↦ \frac{1}{X})$? $f = X+\frac{1}{X}$ would be nice, but I don't know how to check if it's in the given (sub)field.
(End game / context: find $f ∈ \mathbb{Q}(X)$ s.t. $\mathbb{Q}(f) = \mathbb{Q}\left(X+\frac{1}{1-X} + \frac{X-1}{X}\right)^{(X ↦ \frac{1}{X})}$.)
Letting $$g(x)=x+\frac{1}{1-x} + \frac{x-1}{x}$$ it's easily verified that $$g(x)+g\!\left({\small{\frac{1}{x}}}\right)=3$$ hence letting $h(x)=g(x)\bigl(3-g(x)\bigr)$, we get $$h(x)=g(x)\,g\!\left({\small{\frac{1}{x}}}\right)$$ so $h$ satisfies the required conditions.