Are the solutions to the equation $f(n \cdot x)=x$ always expressible in closed form?

Question

Are the solutions to the equation:

$$f(n \cdot x)=x$$

always expressible in closed form?

$$n=1,2,3,4,5,...$$

Answer 1

"Closed form" means i.a. that the set of the allowed functions is given.

$$f(n\cdot x)=x\ \ \ \ \ (n\in\mathbb{N}_+)\tag{1}$$

There are addition theorems for some kinds of functions $f$, e.g. for the exponential functions, the trigonometric functions, the hyperbolic functions, the spherical harmonics and the elliptic functions.

Maybe this is interesting:
Ritt, J. F.: Periodic functions with a multiplication theorem. Trans. Amer. Math. Soc. 23 (1922) (1) 16-25

If $f$ is an algebraic function, Galois theory says which kinds of equation 1 have solutions that are radicals (or elementary numbers). But all solutions of equation 1 can be represented by special functions like elliptic functions, Bring radicals and/or theta functions then. see e.g.:
closed-form expression for roots of a polynomial
How do you solve 5th degree polynomials?

If $f$ is a transcendental function, equation 1 can have algebraic solutions only at the exceptional points of $f(n\cdot x)$.
see e.g. On the behavior of transcendental functions

If $f(n\cdot x)$ is a transcendental expression of $x$ over a non-discrete multi-element domain, $f(n\cdot x)-x$ doesn't have a partial inverse over a non-discrete domain that is an elementary function.

If $f=\exp$, equation 1 doesn't have solutions that are elementary numbers, but it has solutions in terms of Lambert W.

see e.g.:
Example equation which does not have a closed-form solution
Is there an algorithm to determine if a closed-form solution exists?

Answer 2

Depends what you mean by closed form, but if you mean in terms of elementary functions, then no. Consider, for instance, $f(x) = e^x-2$. There is no closed form for $n=1$ much less the other values of $n$.