The existence of such functions being established by now, there remains a curiosity for the concrete situations that may generate them. Here is one, which is also interesting as an example of a $C^\infty$, non-analytic solution, of a simple first order linear delay differential equation.
Let's consider the problem
$$f'(x)=\lambda f(3x),\qquad x < \frac{1}{2} $$ where we seek for a function $f\in C^1(\mathbb{R})$ with support in $I:=[0,1]$, and with the condition $f(x)=f(1-x)$.
I will show that it has a simple eigenvalue $\lambda= \frac{9}{2}$, corresponding to an eigenfunction with $f(\frac{1}{2})=1$ (see the details below). Then, it is very easy to check that $f$ is locally polynomial on the complement of the Cantor set; and it is, of course, $C^\infty$.
Indeed, since $f (3x)=0$ for $ x > \frac{1}{3}$, from the equation we have that $f'(x)$ vanishes identically on the interval $(\frac{1}{3},\frac{2}{3})$, so that $f$ is the constant $1$ therein. By induction, it is easy to see that $f$ is polynomial of degree $k-1$ on each component of the complement of the closed set $F_k$ in $I$, inductively defined by $F_0:=I$ and $F_{k+1}:= \frac{1}{3}F_k \cup\left(\frac{1}{3}F_k + \frac{2}{3}\right)$. Note that $\{F_k\} _ {k\in\mathbb{N}}$ is the well-known nested sequence of closed sets whose intersection is the Cantor set.
$$*$$ Construction. Consider the linear operator $T$ on $L^\infty(0,1)$ (thought as subspace of $L^\infty(\mathbb{R})$ via extension by zero), defined by
$$Tu(x)= \cases{ \int_0^{3x}u(t)\, dt\qquad \qquad 0 < x < \frac{1}{3}\\\ 0 \qquad \qquad \qquad \qquad \frac{1}{3} < x < \frac{2}{3}\\\ \int_0^{3(1-x)}u(t)\, dt \qquad\frac{2}{3}< x < 1.} $$ Since $T$ and $T ^ 2$ are integral operators with non-negative kernels, their norms are attained on the constant function $1$ on $[0,1]$. We have, by simple computations $T1(x)=3\min(x,1-x)\chi _ { F _ 1 }(x) $, $\|T\| = \|T1\| _ \infty=1$ and $\|T^2\|=\|T(T1)\|_\infty=\frac{1}{3}$. Therefore $\rho(T)\le\frac{1}{\sqrt 3} < \frac{2}{3}$ and $I-\frac{3}{2}T$ is invertible. The function
$$f:= \left(I-\frac{3}{2}T\right) ^ {-1}\chi _ {[\frac{1}{3},\frac{2}{3}]} = \sum _ {k=0} ^ \infty \left( \frac{3}{2}\right) ^ k T ^ k \chi _ { [\frac{1}{3}, \frac{2}{3}] } $$
verifies
$$f(x)=\frac{3}{2} \int_0^{3x}f(t)\; dt\qquad \mathrm{for\; all}\quad 0 < x <
\frac{1}{3}\, ,$$
$$f(x)=1\qquad \mathrm{for\, all}\quad \frac{1}{3} < x < \frac{2}{3}\, ,$$
and $$f(x)=f(1-x)\qquad \mathrm{for\, all}\quad x\in\mathbb{R} \, .$$ The only points in $\mathbb{R}$ where the continuity of $f$ is not immediate and has to be checked, are $\frac{1}{3}$ and $\frac{2}{3}$. We have
$$\int_ 0 ^ 1 f(x)dx=\int _ 0 ^ \frac{1}{3} f(x)dx+\int _ \frac{1}{3} ^ \frac{2}{3} f(x)dx+\int _ \frac{2}{3} ^ 1 f(x)dx= 2\int _ 0 ^ \frac{1}{3} f(x)dx+ \frac{1}{3}\, ,$$
and
$$ 2 \int _ 0 ^ \frac{1}{3} f(x)dx = 3\int _ 0 ^ \frac{1}{3} \int _ 0 ^ {3x} f(t) dt dx= \int _ 0 ^ 1 \int _ 0 ^ x f(t) dt dx=$$
$$= \int _ 0 ^ 1 (1-t) f(t) dt = \int _ 0 ^ 1 tf(1-t)dt= \int _ 0 ^ 1 tf(t)dt = $$
$$ =\frac{1}{2}\left(\int _ 0 ^ 1 (1-t)f(t)+ tf(t)dt\right) =\frac{1}{2} \int _ 0 ^ 1 f(t)dt\, . $$
So
$$ \int _ 0 ^ 1f(t)dt = \frac{2}{3}\, ,$$
which implies the continuity of $f$ at $ \frac{1}{3}$ (and by symmetry at $
\frac{2}{3}$ ), since
$$f(\frac{1}{3}-)=\frac{3}{2}\int_ 0 ^ 1 f(x)dx =1= f(\frac{1}{3}+) \, . $$
As a consequence, $f$ actually satisfies
$$f(x)=\frac{3}{2} \int_0^{3x}f(t)\, dt $$
for all $ x < \frac{2}{3}\, ,$
and we conclude that $f$ is in $ C ^ 1(\mathbb{R})$, and has the stated properties.
Here is a Pudding Function $f$, and its integral function, $F(x):=\int_0^x f(t)dt$. Note how the latter coincides with the first third of the former, up to rescaling. Simple considerations based on odd and even symmetry show that $F(1)=\frac{2}{3}$.
Edit 2021 I learned from this answer by Gerald Edgar that this a function quite similar (with a factor 2but not easily related) to this is known since 1966 as the Fabius function. It verifies $F'(x)=2F(2x)$ instead of $f'(x)=\frac{9}{2}f(3x)$.
