Let $A$ be a symmetric $n \times n$ matrix. If $P$ is an $n \times n$ permutation matrix, then $A$ is completely positive if and only if $P^TAP$ is completely positive.
Geometrically is the statement quite obvious. When checking whether $v_1,...v_n$ can be embedded in a nonnegative orthant of some $R^k$, neither the order of the vectors not their length meter.
How can I prove this rigorously?
If $A$ is completely positive then there exists an $m \times n$ matrix $B$ with nonnegative elements such that $A = B^TB$.
For any permutation matrix $P$ we have $$P^TAP = P^TB^TBP = (BP)^T(BP)$$ and $BP$ has nonnegative elements.