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?

One way to think about this is that you are assuming that $f(x) = cx$ and then solving for the value of $c$.  
But there is something much more interesting going on; you have started doing some abstract algebra without knowing it.  $f$ is not a real number, but it does live in something called an algebra over a field, which is a special type of ring.  In particular, $f$ lives in the algebra of continuous functions $\mathbb{R} \to \mathbb{R}$.  In this algebra there is a multiplicative identity $f(x) = x$ which plays the role of the zeroth power, an addition, a multiplication, and a scalar multiplication by real numbers.
Many manipulations which are possible with ordinary real numbers are possible in rings and algebras; in particular, the first half of the proof of the quadratic formula carries through totally abstractly (the part where you complete the square).  
Unfortunately, the second half does not.  In other words, it is not true that there are only two solutions to the equation $f^2 = a$ in a general algebra.  This is because algebras are not in general integral domains.  There may be none or infinitely many!
However, in this special case $a$ is a positive real multiple of the identity, so we know it has at least two square roots (even though there may be more).  These are the solutions that you found, and this method for finding them is perfectly valid. 
This is a very important technique.  It is often used in the case where $f$ is a differential operator as a concise way to solve linear homogeneous ODEs.
