 Solution

For the specified unit vector,For a cyclic matrix with first row $(a,b,c)$, its determinant is $a^3+b^3+c^3-3abc$. Here it givesTherefore $Ax=Cx$ has only the zero solution if $\alpha \ne 1$. At $\alpha =1$, $A=I$ and $A-C$ has kernel consisting of the equal-coordinate vectors. The complete answer isGeometrically, $C^{T}C=I$, $\det C=1$ and $Cn=n$, so $C$ is a rotation matrix about the $n$ axis. Its matrix trace is two, giving $\cos \theta =(\mathrm{tr}C-1)/2=1/2$; its skew part selects angle $-\pi /3$ about the oriented axis $n$. It has no fixed nonzero perpendicular vector. Meanwhile $A$ preserves perpendicular components and scales the axial component by $\alpha$. Comparing these components first forces $x_{\perp }=0$, and then $(\alpha -1)x_{\parallel }=0$, exactly matching the determinant calculation, including the noninvertible case $\alpha =0$.