Step 1: Recall the definition of an eigenvector.
If
\[
A\mathbf{x}=\lambda\mathbf{x},
\]
then
\[
\mathbf{x}
\]
is an eigenvector corresponding to the eigenvalue
\[
\lambda.
\]
Step 2: Apply the scaling property.
If
\[
\mathbf{x}
\]
is an eigenvector, then for any non-zero scalar
\[
k,
\]
\[
A(k\mathbf{x})
=
kA\mathbf{x}
=
k\lambda\mathbf{x}
=
\lambda(k\mathbf{x}).
\]
Hence,
\[
k\mathbf{x}
\]
is also an eigenvector corresponding to the same eigenvalue.
Therefore,
\[
\boxed{
\text{Eigenvectors are unique only up to a non-zero scalar multiple.}
}
\]
Thus,
\[
\boxed{(C)}
\]
is the correct answer.