= Solution
The quadratic expression is $\mathbf x^TA\mathbf x$. Normalize the preceding <eigenvectors> to obtain an <orthonormal basis>
$$
\mathbf u_1=\frac{(1,-1,-1)^T}{\sqrt3},\quad
\mathbf u_2=\frac{(0,1,-1)^T}{\sqrt2},\quad
\mathbf u_3=\frac{(2,1,1)^T}{\sqrt6}.
$$
With the <orthogonal matrix> $U=(\mathbf u_1\ \mathbf u_2\ \mathbf u_3)$ and coordinates $\mathbf y=U^T\mathbf x$, the surface equation becomes $y_1^2+2y_2^2+4y_3^2=1$. All coefficients are positive, so \b[it is an <ellipsoid>], with principal semiaxes $1,1/\sqrt2,1/2$ along the respective <eigenvectors>. This is <principal-axis reduction of a quadric>.
A map to the unit <sphere> is $\mathbf x\mapsto\operatorname{diag}(1,\sqrt2,2)U^T\mathbf x$. One explicit matrix is
$$
\boxed{B=\begin{pmatrix}1/\sqrt3&-1/\sqrt3&-1/\sqrt3\\0&1&-1\\4/\sqrt6&2/\sqrt6&2/\sqrt6\end{pmatrix}.}
$$
Indeed $B^TB=A$, so $|B\mathbf x|^2=\mathbf x^TA\mathbf x$. Since $B$ is invertible, every point on the unit <sphere> has a preimage on the <ellipsoid>; the map is onto, not merely into. The requested $B$ is not unique, since an additional <orthogonal matrix> on the left preserves the unit <sphere>.
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