I made use of the following intermediate step in a demonstration I did: $$...=a_i\sigma_{kj}b_k=a_i\sigma^T_{jk}b_k=...$$ where the "T" just shows that the indices have been flipped.
After looking back on it, I'm not sure why, and if, what I did is true. How did the equality hold after flipping the indices? And wouldn't it (wrongfully) imply that: $$\sigma=\sigma^T$$
In case it's needed, I wrote the whole demonstration below, where I show that: $$(\vec{a}\otimes \vec{b})\bar{\bar{\sigma}}=\vec{a}\otimes (\bar{\bar{\sigma}}^{T}\vec{b})$$
$$((\vec{a}\otimes \vec{b})\bar{\bar{\sigma}})_{ij}=(\vec{a}\otimes \vec{b})_{ik} \sigma_{kj}=a_i b_k \sigma_{kj}=a_i\sigma_{kj}b_k=a_i\sigma^T_{jk}b_k=a_i(\bar{\bar{\sigma}}^{T}\vec{b})_j=(\vec{a}\otimes (\bar{\bar{\sigma}}^{T}\vec{b}))_{ij}$$