Are conjugate vectors unique?

Take $A=I_n$. Then any orthogonal basis is a set of conjugate vectors.

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 $B=DO^T$.

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.