Let's say we have a (3,3) mixed tensor $A^{ijk}_{mnp}= u^i_m v^j_n w^k_p.$
Applying the tensor contraction operation and taking k = p, we have
$A^{ijk}_{mnk}= B^{ij}_{mn}$, a (2,2) mixed tensor.
In case the placement of the equal indices is changed and we want to contract the tensor $A^{ijk}_{knp}$ (setting k = m), do we obtain by contraction the same (2,2) tensor as above? Or do we get another (different) (2,2) tensor of the form
$A^{ijk}_{knp}= C^{ij}_{np}$ ?
You get a different $(2, 2)$ tensor in general. For example, consider the case where $u^i_m = \delta^i_m$, $v^j_n = \delta^j_n$, and $w^k_p = \delta^k_p$. Then $B^{ij}_{mn} = \delta^i_m\delta^j_n\delta^k_k = d\,\delta^i_m\delta^j_n$ where $d$ is the dimension of the vector space, while $C^{ij}_{np} = \delta^i_k\delta^j_n\delta^k_p = \delta^i_p\delta^j_n$.
Viewing tensors as multilinear maps, note that $u, v, w \in \operatorname{End}(V)$ and $A = u\otimes v\otimes w$. Then $B = \operatorname{tr}(w)\, u\otimes v$ and $C = (u\circ w)\otimes v$.