= Solution
For the specified unit <vector>,
$$
A-C=\frac13\begin{pmatrix}
\alpha&\alpha-3&\alpha\\
\alpha&\alpha&\alpha-3\\
\alpha-3&\alpha&\alpha
\end{pmatrix}.
$$
For a cyclic <matrix> with first row $(a,b,c)$, its <determinant> is $a^3+b^3+c^3-3abc$. Here it gives
$$
\det(A-C)=\frac{2\alpha^3+(\alpha-3)^3-3\alpha^2(\alpha-3)}{27}=\alpha-1.
$$
Therefore $Ax=Cx$ has only the zero solution if $\alpha\ne1$. At $\alpha=1$, $A=I$ and $A-C$ has kernel consisting of the equal-coordinate <vectors>. The complete answer is
$$
\boxed{\alpha\ne1:\ x=0;\qquad \alpha=1:\ x=t(1,1,1),\quad t\in\mathbb R.}
$$
Geometrically, $C^TC=I$, $\det C=1$ and $Cn=n$, so $C$ is a <rotation matrix> about the $n$ axis. Its <matrix trace> is two, giving $\cos\theta=(\operatorname{tr}C-1)/2=1/2$; 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 $\alpha=0$.
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