Why spherical harmonics are related to certain rotations (and not others)? Let's take a direction eigenket $|{\bf\hat{n}}\rangle$ in 3-dimensional space oriented with angles $\theta\in\left[0,\pi\right]$ and $\phi\in\left[0,2\pi\right]$ in spherical coordinates. Next take the $|{\bf\hat{z}}\rangle$ direction eigenket. 
There are infinite rotations which take $|{\bf\hat{z}}\rangle$ to $|{\bf\hat{n}}\rangle$, but in particular we can consider a specific rotation given by the following operator (Wigner D-Matrix) written with Euler angles $(\alpha,\beta,\gamma)$:
\begin{align}
&\mathcal{D}(R)=\mathcal{D}(\alpha=\phi,\beta=\theta,\gamma=0)\\
&\implies|{\bf\hat{n}}\rangle=\mathcal{D}(R)|{\bf\hat{z}}\rangle
\end{align}
Notice that if we change the value of $\gamma$ the equation remains valid (because $\gamma$ represents a rotation about ${\bf \hat{n}}$ axis). This is the same notation that appears on Sakurai and Wikipedia. Next it is shown a relation with spherical harmonics (from Sakurai & Napolitano, sec. 3.6 pp 205-206):
\begin{align}
&|{\bf\hat{n}}\rangle=\mathcal{D}(R)|{\bf\hat{z}}\rangle\\
\implies &|{\bf\hat{n}}\rangle=\sum_{l',m'}\mathcal{D}(R)|l',m'\rangle\langle l',m'|{\bf\hat{z}}\rangle\\
\implies \langle l,m &|{\bf\hat{n}}\rangle=\langle l,m|\sum_{l',m'}\mathcal{D}(R)|l',m'\rangle\langle l',m'|{\bf\hat{z}}\rangle\\
\text{Using that}&\ \text{$\mathcal{D}(R)$ doesn't change the $l'$ number:}\\
\implies Y_{l}^{m*}(\theta,\phi)&=\sum_{m'}\langle l,m|\mathcal{D}(R)|l,m'\rangle\langle l,m'|{\bf\hat{z}}\rangle\\
\implies Y_{l}^{m*}(\theta,\phi)&=\sum_{m'}\mathcal{D}^{(l)}_{m,m'}(R) Y_l^{m'}(\theta=0,\phi)\\
\text{And using that}&\ \text{$Y_l^{m'}$ vanishes at $\theta=0$ for $m'\neq 0$:}\\
\implies Y_{l}^{m*}(\theta,\phi)&=\sqrt{\frac{2l+1}{4\pi}}\mathcal{D}^{(l)}_{m,0}(R)\\
\end{align}
This gives a relation between the rotation operator $\mathcal{D}(R)$ and spherical harmonics. It does not seem in any step of the derivation that $\gamma=0$ is necessary. I think that if we choose another rotation $R$ with $\gamma\neq 0$ such that $|{\bf\hat{n}}\rangle=\mathcal{D}(R)|{\bf\hat{z}}\rangle$ the relation remains valid, but this is weird because if $R$ changes then the operator $\mathcal{D}(R)$ changes (and $Y_l^{m*}$ doesn't change). 
My question is: Why the spherical harmonics represent rotations with $\gamma=0$ and not another value? and, if we have a rotation with $\gamma\neq 0$, can we write it with spherical harmonics?
 A: The simplest explanation starts by writing explicitly the first Euler rotation about the $\boldsymbol{\hat z}$ as 
$$
R_z(\gamma)=
\left(\begin{array}{ccc}
\cos\gamma&\sin\gamma & 0 \\
-\sin\gamma &\cos\gamma &0 \\
0&0&1
\end{array}
\right)
$$
Clearly 
$$
R_z(\gamma)\hat z = R_z(\gamma) \left(\begin{array}{c} 0 \\ 0 \\ 1\end{array}\right)= \left(\begin{array}{c}0 \\ 0 \\ 1\end{array}\right)
$$
independent of the angle $\gamma$.  Thus, your function must also be independent of $\gamma$; this is achieved by setting $\gamma=0$.  
[By the way this rotation by $\gamma$ is about $\boldsymbol{\hat z}$, not ${\boldsymbol{\hat n}}$.]
A: The other $D^{(j)}_{mn}(\theta, \phi \psi) $ with $n\ne 0$ play the role  of
monopole harmonics i.e the angular part of the wavefunctions of a charged particle moving around a magnetic monopole.     The coordinate set $\theta, \phi, \psi $ parameterize the three sphere $S^3$ and the Hopf map is ${\rm Hopf}: (\theta, \phi, \psi)\mapsto (\theta, \phi)\in S^2$ and the Monopole harmonics are section of the fibre bundle whose total space is $S^3$ and the Hopf map is the projection to the base space.
As functions on $S^2$  one has to make a choice of the third Euler angle $\psi$ at each direction $\theta$ $\phi$. This is a gauge choice
