integration - I-linear functions

I am trying to understand how many functions $f:\mathbb R\to \mathbb R$ such that
$$\int f(g(x))\mathrm d x=f\left ( \int g(x)\mathrm d x \right) \tag{$\star$}$$
for every $g:\mathbb R \to \mathbb R$ exist. We shall refer to a function satisfying $(\star)$ as an I-linear function.

It is easy to show that the identity function is I-linear. Moreover, the space of I-linear functions forms a vector space because if $f_1,f_2$ are I-linear
$$ \begin{align}(\alpha f_1+\beta f_2) \left (\int g(x)\mathrm d x \right ) &= \alpha f_1 \left (\int g(x)\mathrm d x \right)+\beta f_2 \left ( \int g(x)\mathrm d x \right)
\\
&= \alpha \int f_1(g(x))\mathrm d x+\beta \int f_2(g(x))\mathrm d x\\
&= \int (\alpha f_1+\beta f_2)(g(x))\mathrm d x
\end{align}$$

As a consequence, all scalar multiples of identity are I-linear functions. This claim is essentially equivalent to stating the obvious $\int \alpha f=\alpha \int f$.

Observe that if $f$ is both I-linear and differentiable, by setting $g(x)=e^x$ and differentiating $(\star)$ one obtains
$$f(e^x)=e^xf'(e^x+C)=\frac{\mathrm d}{\mathrm d x}f(e^x+C)$$
which looks almost like a differential equation.

Questions:

1. Under no regularity hypotheses on $f$, is it possible to show that if $f$ is I-linear, then $f(x)=\alpha x$ for some $\alpha$? To do so, it would be sufficient to show that the vector space of I-linear functions is $1$-dimensional.


2. If the claim is false under no regularity hypotheses, is it possible to add these to prove our claim?