When does $x^n = \sqrt[n]x$ for odd $n$ and real $x$?

Lead-up to the problem

Given that $n \in \mathbb Z$ and $x \in \mathbb R$, the function $f(x) = x^n$ will have an inverse only for odd values of $n$.

We find the inverse in these cases to be $f^{-1}(x) = g(x) = \sqrt[n]x$.

Problem

Find the points of intersection between $f$ and its inverse $g$.

Thoughts

Essentially we have to solve the equation $x^n = \sqrt[n]x$ for odd $n$ and real $x$.

Doing any "real" calculations gives arbitrarily powered equations like $x^{n^2} - x = 0$ but beyond the trivial solutions $x=0, 1$ or $n = \pm 1$, it's hard to predict if it's even worth trying to find more.

Conclusion (?)

$x \in \{0, 1\} \ \vee \ n=\pm1$ must be the only solutions.

Does that look right?

Solution 1:

You can see that $|x^m| > |x|$ for all $m >1$ and all $x$ satisfying $|x| >1$. So $x^{n^2} \not = x$ for all $x$ satsifying $|x| >1$. Thus, the aforementioned solutions $x \in \pm 1,0$ are precisely the set of solutions.

ETA: For whatever reason I had read $n$ must be a positive integer. The same line of reasoning holds for every odd $n \not \in \pm 1$ positive or negative, except $x$ cannot be $0$ for negative $n$. For $n=1,-1$, note that $x^n =x^{1/n}$ for all $x \not = 0$; for $n=-1$, and for all $x$ for $n=1$.