Are conjugate vectors unique? A set of nonzero vectors $\{p_0,p_1,\ldots,p_{n-1}\}$ is said to be conjugate with respect to a symmetric positive definite matrix $A$ if
$$
p_i^{\mathrm T}Ap_j=0
$$
for all $i\ne j$. Such vectors are used in the conjugate gradient method. It follows that conjugate vectors are also linearly independent. An example of conjugate vectors are the eigenvectors of the matrix $A$.

Are the conjugate vectors unique up to a scalar multiple? In other words, if there are two sets of conjugate vectors with respect to the same symmetric positive definite matrix $A$, are the vectors going to be the same up to a scalar multiple?

I would guess that that they are unique up to a scalar multiple, but I am not sure.
Any help is much appreciated!
 A: No, they are far from unique.
Let $P$ be the matrix whose columns are the vectors $p_i$. Then, the vectors form a conjugate family with respect to a symmetric positive definite matrix $A$ if and only if $P^TAP=D$ is a diagonal matrix. You are asking if $P$ is unique up to scalar multiple, i.e. multiplying on the right by a diagonal matrix. If you enforce that the diagonal of $D$ is non-zero, then you can consider $P'=PD^{-1/2}$, which satisfies $P'^TAP'=I$, the identity matrix. Now replacing $P'$ with $P'O$ for any orthogonal matrix $O$ results in $(P'O)^TA(P'O)=I$, forming another conjugate family from the one we started with. If instead there are some zeros on the diagonal, you get even more degeneracy: the matrix $D$ can be block decomposed into a block non-zero diagonal entries, and the rest a block of zeros. Then you can perform the same procedure mentioned above on the non-zero block, and apply any transformation whatsoever to the zero portion.
A: Take $A=I_n$. Then any orthogonal basis is a set of conjugate vectors. 
A: In general, no. If $A=B^TB$ for some square matrix $B$, $p_i^TAp_j=(Bp_i)^T(Bp_j)$. Since such a $B$ is invertible, any orthogonal basis $e_i$ of the original vector space gives a choice of the $p_i$ as $B^{-1}e_i$. In fact, a diagonalisation $A=OD^2O^T$ will allow us to choose $B=DO^T$.
