The equation $ 2^x = x^2 $ has two real obvious solutions $(x=2,4)$ and another root not so obvious $x\approx- 0.766665$;
Assuming existence and uniqueness, extending equality $f(x)= g(x)$ between such exponentiation swap to real functions $f(x), g(x)$ as well, in
$$ f(x)^ {g(x)}= g(x)^ {f(x)}$$
what other non-obvious real continuous function solutions may be found (implicitly or explicitly) ? Like e.g., $ F(f(x),g(x))=0 $, or ODEs that could redefine exponential or log functions?
Thanks for all related further directions. I could not get beyond
$$ \frac{f(x)}{g(x)}=\frac{\log f(x)}{\log g(x)} $$
Let $h(x)=\dfrac{\log x}{x}$. The range of this function is $(-\infty,e^{-1}]$. For $c\le0$, and for $c=e^{-1}$, there is a unique $x$ (in $(0,1]$ in the first case and $e$ in the second) for which $h(x) = c$, but for $0<c <e^{-1}$, there are two values of $x$, one in $(1, e)$ and the other in $(e, \infty)$, for which $h(x)=c$. Let $\phi$ be the function defined on $(0,\infty)$ that maps one of these values to the other for arguments greater than 1, maps arguments less than or equal to 1 to themselves and maps $e$ to $e$.
Your question can be written as $h(f(x)) = h(g(x))$, so is solved by choosing any function $f(x)$ with range a subset of $(0,\infty)$ and letting $g(x)=f(x)$ or $\phi(f(x))$ for each $x$. There are restrictions on your choice if you want the functions to be 'nice' - for instance, continuous.