Solving (quadratic) equations of iterated functions, such as $f(f(x))=f(x)+x$ In this thread, the question was to find a $f: \mathbb{R} \to \mathbb{R}$ such that
$$f(f(x)) = f(x) + x$$
(which was revealed in the comments to be solved by $f(x) = \varphi x$ where $\varphi$ is the golden ratio $\frac{1+\sqrt{5}}{2}$).

Having read about iterated functions shortly before though, I came up with this train of thought:
$$f(f(x)) = f(x) + x$$
$$\Leftrightarrow f^2 = f^1 + f^0$$
$$f^2 - f - f^0 = 0$$
where $f^n$ denotes the $n$'th iterate of $f$.
Now I solved the resulting quadratic equation much as I did with plain numbers
$$f = \frac{1}{2} \pm \sqrt{\frac{1}{4} + 1}$$
$$f = \frac{1 \pm \sqrt{1+4}}{2} = \frac{1 \pm \sqrt{5}}{2}\cdot f^0$$
And finally the solution
$$f(x) = \frac{1 \pm \sqrt{5}}{2} x .$$
Now my question is: *Is it somehow allowed to work with functions in that way?** I know that in the above, there are denotational ambiguities as $1$ is actually treated as $f^0 = id$ ... But since the result is correct, there seems to be some correct thing in this approach.
So can I actually solve certain functional equations like this? And if true, how would the correct notation of the above be?

Whether or not $f$ is linear, the equation can in fact be rewritten as $F^2 = F + 1$, in a suitable interpretation where to a function $f$ is associated a linear operator $F$.
Let $V$ be the vector space of sums $aX + bf(X)$. Denote by $F$ the operator that takes an element of $V$ to its composition with $f$, that is,  $F (h(x)) = h(f(x))$.  Let $1$ denote the identity operator on $V$.
Then $F$ is linear (from the definition), and satisfies $F^2=F+1$ as one can check on basis elements.
I have been temporarily but deliberately unclear as to whether $V$ is the 2-dimensional space of formal sums with basis vectors $x$ and $f(x)$, or the at most 2-dimensional space of functions of that form.   The question of whether $f$ is linear is the same as asking whether the second space is one-dimensional, which is no longer an algebraic question but one of analysis using regularity assumptions on $f$. (In particular, seeing $f$ as an operator or 2x2 matrix $F$, illuminates but does not quite trivialize the problem of showing that linear functions are the only continuous solutions to the functional equation. Additional arguments are needed.)   The equation $F^2 = F + 1$ holds in both interpretations.
