If there is such a function, it can't be represented by Tailor series:
Given: $f( \left(\int_{0}^{t} g(x) \ \text{d}x\right)) = g( \left(\int_{0}^{t} f(x) \ \text{d}x\right))$
$f( G(t) $ -$ G(0)$ ) = g( F(t) -$ F(0)$ )
$g(x)=g_0+g_1x+g_2x^2+...$
$f(x)=f_0+f_1x+f_2x^2+...$
$G(x)=g_0x+g_1\frac{x^2}2+g_2\frac{x^3}3+...$
$F(x)=f_0x+f_1\frac{x^2}2+f_2\frac{x^3}3+...$
$f_0+f_1(g_0x+...)+f_2(g_0x+...)^2+...=g_0+g_1(f_0x+...)+g_2(f_0x+...)^2+...$
Since there's only one term of $kx^0$ and $kx^1$ on each side, we can conclude their respective constants are equal:
$f_0=g_0,f_1=g_1$
For every power afterward it is possible to substitute every previous set of constants in order to obtain another.
$f_2=g_2, f_3=g_3, ...$ (Hence $f(x)=g(x)$)