BTW, just to make your head spin, I did the same calculation in octave. I get:
octave:1> A = [[-1,10]; [-4,12]]
octave:3> [V, LAMBDA] = eig(A)
So I get the first one that MS shows and a new third one to add to your collection. Matlab produces the same result, BTW.
As I said, the eigenvector appears on both sides of the equation, so it is not "unique" in the sense that you are constrained to just n-dof unique vectors that solve the equation. There are a multiplicity of solutions that are linearly dependent/scalable. What you end up with will depend upon your method of finding them.
The eigenvalues, however, are unique. The appear only on one side of the equation. So they cannot be scaled. That makes sense in that the eigenvalues are the roots of the characteristic polynomial, so they cannot be scaled like the eigenvectors.