The only $V$-invariant probability measure is the delta concentrated on the origin. A quick way to see it is to look at the conjugate of $V$ with the multiplication operator $M$ by $e^x$. Indeed we have for any $f\in C_0$ and $x\in[0,1]$

$$\big|M^{-1}VMf(x)\big|=\Big|e^{-x}\int_0^{\sqrt x}f(s)e^s ds\Big|\le e^{-x}\int_0^{\sqrt x}e^s ds\|f\|_\infty =$$$$=e^{-x}\big(e^{\sqrt x}-1\big)\|f\|_\infty \le{2\over3}\|f\|_\infty.$$

So $\|M^{-1}VM\|\le 2/3$. This implies that $\|M^{-1}V^nM\|\le (2/3)^n$ for any $n$, and therefore for any $r>0$ we have an inclusion
$V^n(B(0,r))\subset V^nM(B(0,r))\subset B(0, ({2/3})^ner)$, and finally
$$B(0,r)\subset V^{-n}B(0, ({2/3})^ner).$$
Being $r$ and $n$ arbitrary, this implies that any $V$-invariant probability measure $\mu$ gives the same value to any nbd of $0$.

As a general principle, the same conclusion hold for any bounded operator with spectral radius less than $1$: up to conjugation it is a norm contraction, and the same argument applies.