Let us consider $\mathbb{R}^3$. We want to rotate a vector $n$ by a magnitude $\alpha$ around a vector $z$ and afterwards by $\beta$ around a vector $z^\perp$. For simplicity, all vectors are unit vectors, pointing away from the origin $(0,0,0)$. The vector $z$ is not $z=(0,0,1)$ but any unit vector. The vector $z^\perp$ is some vector that is perpendicular to $z$.
We can compute the matrix $R_{z,\alpha}$ such that $R_{z,\alpha} n $ describes the rotation of $n$ around $z$ by $\alpha$. Similarly we define $R_{z^\perp , \, \beta}$.
Obviously $$ R_{z^\perp , \, \beta} R_{z,\alpha} n\not = R_{z,\alpha} R_{z^\perp , \, \beta} n$$ in general. But can we change the angle $\beta$, such that $$ R_{z^\perp , \, \beta} R_{z,\alpha} n = R_{z,\alpha} R_{z^\perp , \, \hat{\beta}} n$$ for some $\hat{\beta}$? If so, how does $\hat{\beta}$ relate to $\alpha$, $\beta$, $z$, and $z^\perp$.?
$$R_{z^\perp , \, \beta} R_{z,\alpha} = R_{z,\alpha} R_{z^\perp , \, \hat{\beta}} \\ \text { if and only if} \\ R^{-1}_{z,\alpha} R_{z^\perp , \, \beta} R_{z,\alpha} = R_{z^\perp , \, \hat{\beta}} . $$
Now $R^{-1}_{z,\alpha} R_{z^\perp , \, \beta} R_{z,\alpha}$ is a rotation with axis $R^{-1}_{z,\alpha}z^\perp$ through an angle $\beta$. So what you want is not usually possible.