Does equation $f(g(x))=a f(x)$ have a solution? Suppose $g: [0,1] \to [0,1]$ is a strictly increasing, continuously differentiable function with $g(0)=0$ and $g(1) < 1$, and $a \in (0,1)$.
Is there a function $f:[0,1] \to \mathbb{R}$ such that $f(g(x))=a f(x)$ for all $x \in [0,1]$?
I know that when $g$ is linear, then there are many such functions $f$. But how about arbitrary (sufficiently smooth) function?
Background: it is a part of larger project and I have already shown that $g$ is a certain function with nice properties. I have strong intuition that $f$ function should exists. If not generally, then with some additional properties that $g$ probably has. Also, if it helps, then in all cases relevant to my project, $a$ and $g(1)$ are small.
 A: Theorem For each $a\in (0,1)$ there are infinitely many $f\in C([0,1])$ so that $f(0)=0$, $f(1) = 1$ and 
$$(*)\ \ f (g(x) ) = af(x)\ \ \ \  \forall x\in [0,1].$$
Remark: We set $f(1) = 1$ just as a normalization as $(*)$ is linear, as observed in another answer. 
Proof of theorem: Let 
$$C = \{x\in [0,1]: \ g(x) = x\}.$$
$C$ is nonempty as $0\in C$. Let $c = \sup C$. Then $c <1$ as $g(1)<1$. 
Claim 1: $g(x)<x$ for all $x>c$ and $g(x)<c$ for all $x<c$. 
Proof of claim 1: As $x>c$, $x\notin C$ and thus either $g(x) <x$ or $g(x)>x$. If $g(x)>x$, then the function $h(x) = g(x)>x$ satisfies $h(x)>0$ for some $x<1$ and $h(1)<0$. Thus there is $x'>c$ so that $h(x')=0$, which is impossible by definition of $c$. The second inequality comes from $x<c \Rightarrow g(x)<g(c)=c$. 
Claim 2: Let $I_1 = (g(1), 1]$ and inductively $I_n = g(I_{n-1})$ for $n>1$. Then the $I_i$'s are pairwise disjoint nonempty intervals, $g : I_{n} \to I_{n+1}$ is bijective with continuous inverse, and 
$$(**)\ \ \ \ \bigcup_{n \in \mathbb N} I_n = (c, 1].$$
Proof of claim 2: The claim is quite obvious as $g$ is strictly increasing and $I_{n} = (g^n(1), g^{n-1}(1)]$. To show $(**)$, it suffices to show that $g^n(1) \to c$ as $n\to \infty$. First of all, one can show inductively that $g^n(1) >c$ as 
$$g^n(1) > c\Rightarrow g^{n+1} (1) > g(c) =c$$
Note that the sequence is strictly decreasing. Thus it converges to some $d\in [0,1]$. Then $g(d) = d$ and so $d\in C$. Together with $g^n(1) >c$ we have $d = c$. 
Now we define $f$. 
From $(*)$, as $a\neq 1$, we have $f(x) = 0$ for all $x\in C$. Thus we simply set $f(x) = 0$ on $[0,c]$ (Note that by the second equality of claim 1, $(*)$ is satisfied on $[0,c]$). Now we define $f$ inductively on each $I_n$. First on $I_1$, let $f$ be any continuous function so that $f(1) = 1$, $f(g(1)) = a$. 
Now on $I_2$ we define $f(y) = a f(g^{-1}(y))$. Note that this is well defined as $g^{-1}(y) \in I_1$. This $f$ is continuous on $I_2$ as $g^{-1}$ is continuous. On $g(1) \in I_2$, we have 
$$  f(g(1)) = a f(g^{-1}(g(1)) = af(1)  =a$$
Thus $f$ is continuous on $I_2\cup I_1 = (g^2(1), 1]$. Also by construction, $f$ satisfies $(*)$ on $I_2 \cup I_1$. Note also that $||f||_{I_2} \le a ||f||_{I_1}$ by construction (This is also observed in the comment). 
Now by induction, assume that $f$ has been continuously defined on $I_1\cup \cdots \cup I_n$ so that $(*)$ is satisfied and $||f||_{I_k} = a^{k-1} ||f||_{I_1}$ for all $k =1, 2, \cdots n$. Then define on $I_{n+1}$ 
$$f(y) = a f(g^{-1} (y)).$$
Then similarly one can show that $(*)$ is satisfied and $||f||_{I_{n+1}} \le a^n ||f||_{I_1}$. By induction and claim 2, $f$ has been defined on $[0,1]$ and satisfies $(*)$. The only thing we need to check is that $f$ is continuous at $c$, but it is obvious as $||f||_{I_n} \le a^n ||f||_{I_1}$ and $|a|<1$. This concludes the proof of the theorem. 
Corollary $f$ can be chosen to be nondecreasing. If $g(x)<x$ for all $x>0$, then $f$ can be chosen to be strictly increasing. 
Proof of corollary The first statement is true as we can initially choose $f$ to be strictly increasing on $[g(1), 1]$, as $a<1$. Then $f$ would be strictly increasing on $[c, 1]$. If in addition that $g(x)<x$ for all $x>0$, $C = \{0\}$, so $c = 0$ and thus $f$ is strictly increasing. 
Remark: In some sense, it seems hard to study the spectral theory of the operator $f\mapsto f\circ g$ on $C([0,1])$, as all the eigenspaces are infinite dimensional. Also this is impossible to use fixed point theory as such $f$ are not unique. 
A: you are defining a linear transformation h on  C[0,1] defined by h(f)=f(g) and you are trying to find eigen vectors and values. (a) which you are considering is an eigen value and f is an eigen function. Why not consider analytic functions and then we see that g(x)=cx+dx^2+... while f(x)=A+Bx+Cx^2+.... Then we see that A=0. It seems the problem is solvable in this case. 
