Haar measure of SU(2): Align to the z-direction for class functions

Any element of $\mathrm{SU}(2)$ can be parametrized via $$g=e^{i\varphi\vec{n}\cdot\vec{\sigma}}$$

where $\vec{n}$ is a normal vector and where $\varphi\in [0,2\pi]$. The matrices $\vec{\sigma}=(\sigma_{1},\sigma_{2},\sigma_{3})$ are the Pauli-matrices. With respect to this parametrization, the noramlized Haar measure is given by $$\mathrm{d}g=\frac{1}{\pi}\mathrm{sin}(\varphi)^{2}\mathrm{d}\varphi\mathrm{d}^{3}\vec{n}$$ Now, my question is: Is it always possible to use bi-invariance of the Haar-measure to rotate the group element to the z-direction? In other words, is it always possible to write

$$\int_{\mathrm{SU}(2)}\,\mathrm{d}g\,f(g)=\frac{1}{\pi}\int_{0}^{\pi}\,\mathrm{d}\varphi\,\mathrm{sin}(\varphi)^{2}\,f(e^{i\varphi\sigma_{z}})?$$

where $f$ is a class function, i.e. $f(h)=f(ghg^{-1})$.

Your integral is off by as half, but is otherwise correct for class functions.

Lets review some facts. The element $e^{i\varphi n\sigma}$ has eigenvalues $e^{i\varphi},e^{-i\varphi}$, so namely, they are completely a function of $\varphi$. In addition, there is a conjugation by an element of $SU(2)$ that brings any element of $e^{i\varphi n\sigma}$ to $e^{i\varphi \sigma_z}$ (this follows from explicit computation or basic maximal torus theory).

In particular, we may now write $$\int_{SU(2)} dg f(g)=\int_{SU(2)}dgf(e^{i\varphi \sigma_z})=\frac{1}{\pi}\int_{0}^{2\pi}d\varphi\sin(\varphi)^2\int dn f(e^{i\varphi \sigma_z})=$$$$\frac{1}{\pi}\int_0^{2\pi}d\varphi\sin(\varphi)^2f(e^{i\varphi \sigma_z}).$$

Now a last note. Observe that here the integral goes from $[0,2\pi]$, as is the case for the actual Haar measure decomposition. On the other hand, as $e^{i\varphi \sigma_z}$ is conjugate to $e^{-i\varphi\sigma_z}$, we may write $$\frac{1}{\pi}\int_0^{2\pi}d\varphi\sin(\varphi)^2f(e^{i\varphi \sigma_z})=\frac{2}{\pi}\int_0^{\pi}d\varphi\sin(\varphi)^2f(e^{i\varphi \sigma_z}).$$