Constructing a functional equation that has given solution set.

Motivation was that all the functional equations that appear in Olympiad have mainly one solution. for example,
$xf(x^2)f(f(y))+f(yf(x))=f(xy)(f(f(x^2))+f(f(y^2)))$

or

$f(x+y)^2=2f(x)f(y)+max(f(x^2+y^2),f(x^2)+f(y^2))$

You can find infinitely many of them in AOPS site.

Let's consider following functional equations.

$f(xy)=f(x)f(y)$

this clearly has infinitely many non-trivial solutions, that is, $f(x)=x^a$

My question is, does my following conjecture hold?

Conjecture: we call functional equation GOOD if $f(x)=x$ is a solution, and it is consist of algebraic operation of $f$. If there are finitely many non-constant solutions. Then the equation's only non-trivial solution is $f(x)=x$.

My question could be unclear, so basically I'm asking the following.

Can you make a functional equation that only $f(x)=x, f(x)=x^2$ is solution?

More generally, is it possible to construct a functional equation that the solution set is given set?