First, this belongs to a functional equation of the form http://eqworld.ipmnet.ru/en/solutions/fe/fe2308.pdf
Let $\begin{cases}z_1=\ln z\\V_1(z_1)=V(z)\end{cases}$ ,
Then $V_1(dz_1)=de^{(d-1)z_1}V_1(z_1)+\sum\limits_{k=2}^db_ke^{(d-k)z_1}$
Then, this belongs to a functional equation of the form http://eqworld.ipmnet.ru/en/solutions/fe/fe2303.pdf
Let $\begin{cases}z_2=\ln z_1\\V_2(z_2)=V_1(z_1)\end{cases}$ ,
Then $V_2(z_2+\ln d)=de^{(d-1)e^{z_2}}V_2(z_2)+\sum\limits_{k=2}^db_ke^{(d-k)e^{z_2}}$
$z_2\to z_2\ln d$ :
$V_2(z_2\ln d+\ln d)=de^{(d-1)e^{z_2\ln d}}V_2(z_2\ln d)+\sum\limits_{k=2}^db_ke^{(d-k)e^{z_2\ln d}}$
$V_2((z_2+1)\ln d)=de^{(d-1)d^{z_2}}V_2(z_2\ln d)+\sum\limits_{k=2}^db_ke^{(d-k)d^{z_2}}$
$\therefore V_{2,c}((z_2+1)\ln d)=de^{(d-1)d^{z_2}}V_{2,c}(z_2\ln d)$
$V_{2,c}(z_2\ln d)=\Theta_1(z_2)\prod\limits_{z_2}de^{(d-1)d^{z_2}}$ , where $\Theta_1(z_2)$ is an arbitrary periodic function with unit period
According to http://en.wikipedia.org/wiki/Indefinite_product#Rules,
$V_{2,c}(z_2\ln d)=\Theta_1(z_2)d^{z_2}\left(\prod\limits_{z_2}e^{d^{z_2}}\right)^{d-1}$ , where $\Theta_1(z_2)$ is an arbitrary periodic function with unit period
$V_{2,c}(z_2\ln d)=\Theta_1(z_2)d^{z_2}\left(e^{\sum\limits_{z_2}d^{z_2}}\right)^{d-1}$ , where $\Theta_1(z_2)$ is an arbitrary periodic function with unit period
According to http://en.wikipedia.org/wiki/Indefinite_sum#Antidifferences_of_exponential_functions,
$V_{2,c}(z_2\ln d)=\Theta_1(z_2)d^{z_2}\left(e^{\frac{d^{z_2}}{d-1}}\right)^{d-1}$ , where $\Theta_1(z_2)$ is an arbitrary periodic function with unit period
$V_{2,c}(z_2\ln d)=\Theta_1(z_2)d^{z_2}e^{d^{z_2}}$ , where $\Theta_1(z_2)$ is an arbitrary periodic function with unit period
Let $V_2(z_2\ln d)=d^{z_2}e^{d^{z_2}}W_2(z_2\ln d)$ ,
Then $d^{z_2+1}e^{d^{z_2+1}}W_2((z_2+1)\ln d)=de^{(d-1)d^{z_2}}d^{z_2}e^{d^{z_2}}W_2(z_2\ln d)+\sum\limits_{k=2}^db_ke^{(d-k)d^{z_2}}$
$d^{z_2+1}e^{d^{z_2+1}}W_2((z_2+1)\ln d)=d^{z_2+1}e^{d^{z_2+1}}W_2(z_2\ln d)+\sum\limits_{k=2}^db_ke^{(d-k)d^{z_2}}$
$W_2((z_2+1)\ln d)=W_2(z_2\ln d)+\sum\limits_{k=2}^d\dfrac{b_ke^{-kd^{z_2}}}{d^{z_2+1}}$
$W_2(z_2\ln d)=\Theta_1(z_2)+\sum\limits_{z_2}\sum\limits_{k=2}^d\dfrac{b_ke^{-kd^{z_2}}}{d^{z_2+1}}$ , where $\Theta_1(z_2)$ is an arbitrary periodic function with unit period
$W_2(z_2\ln d)=\Theta_1(z_2)+\sum\limits_{k=2}^d\sum\limits_{z_2}\dfrac{b_ke^{-kd^{z_2}}}{d^{z_2+1}}$ , where $\Theta_1(z_2)$ is an arbitrary periodic function with unit period
According to http://en.wikipedia.org/wiki/Antidifference#Mueller.27s_formula,
$\because\lim\limits_{z_2\to+\infty}\dfrac{b_ke^{-kd^{z_2}}}{d^{z_2+1}}=0$
$\therefore W_2(z_2\ln d)=\Theta_1(z_2)-\sum\limits_{k=2}^d\sum\limits_{n=0}^\infty\dfrac{b_ke^{-kd^{z_2+n}}}{d^{z_2+n+1}}$ , where $\Theta_1(z_2)$ is an arbitrary periodic function with unit period
$V_2(z_2\ln d)=\Theta_1(z_2)d^{z_2}e^{d^{z_2}}-e^{d^{z_2}}\sum\limits_{k=2}^d\sum\limits_{n=0}^\infty\dfrac{b_ke^{-kd^nd^{z_2}}}{d^{n+1}}$ , where $\Theta_1(z_2)$ is an arbitrary periodic function with unit period
$V_2(z_2\ln d)=\Theta_1(z_2)e^{z_2\ln d}e^{e^{z_2\ln d}}-e^{e^{z_2\ln d}}\sum\limits_{k=2}^d\sum\limits_{n=0}^\infty\dfrac{b_ke^{-kd^ne^{z_2\ln d}}}{d^{n+1}}$ , where $\Theta_1(z_2)$ is an arbitrary periodic function with unit period
$V_2(z_2)=\Theta(z_2)e^{z_2}e^{e^{z_2}}-e^{e^{z_2}}\sum\limits_{k=2}^d\sum\limits_{n=0}^\infty\dfrac{b_ke^{-kd^ne^{z_2}}}{d^{n+1}}$ , where $\Theta(z_2)$ is an arbitrary periodic function with period $\ln d$
$V(z)=\Theta(\ln\ln z)z\ln z-z\sum\limits_{k=2}^d\sum\limits_{n=0}^\infty\dfrac{b_ke^{-kd^n\ln z}}{d^{n+1}}$ , where $\Theta(z)$ is an arbitrary periodic function with period $\ln d$
$V(z)=\Theta(\ln\ln z)z\ln z-z\sum\limits_{k=2}^d\sum\limits_{n=0}^\infty\dfrac{b_kz^{-kd^n}}{d^{n+1}}$ , where $\Theta(z)$ is an arbitrary periodic function with period $\ln d$
The solution should be not unique.
So McMullen claims should be wrong. I think he should first make deep introspection about the form of the general solution of the functional equation $V(z^d)=f(z)V(z)+g(z)$ .