= Solution
The first operation acts nearest the column vector, so the composite <rotation matrix> is $M=R_z(\alpha)R_x(\beta)$:
$$
\boxed{M=\begin{pmatrix}
\cos\alpha&-\sin\alpha\cos\beta&\sin\alpha\sin\beta\\
\sin\alpha&\cos\alpha\cos\beta&-\cos\alpha\sin\beta\\
0&\sin\beta&\cos\beta
\end{pmatrix}.}
$$
At $\alpha=\beta=\pi/2$ this reduces to
$$
M=\begin{pmatrix}0&0&1\\1&0&0\\0&1&0\end{pmatrix},\qquad
(x,y,z)\mapsto(z,x,y).
$$
Solving $M\mathbf v=\mathbf v$ gives $v_x=v_y=v_z$. Normalization yields the fixed-axis <unit vectors>
$$
\boxed{\mathbf v=\pm\frac1{\sqrt3}(1,1,1).}
$$
Both signs describe the same unoriented axis. The operation is a cyclic permutation, and $M^3=I$; it is not generally obtained by reversing the multiplication order.
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