$\displaystyle T:V \rightarrow W$ is a linear transformation and $\displaystyle S \in L^k (W).$
Is it true that $\displaystyle T^*(S^{\delta})= (T^* (S))^{\delta}, \delta \in S_k$?
Here is what I did, but unfortunately it was graded as incorrect.
On the k-tensor powers the induced map is $\displaystyle T:V^{\otimes k}\to W^{\otimes k}$ which on pure tensors is $\displaystyle T(v_1\otimes v_2\otimes \ldots \otimes v_k)=T(u_1)\otimes T(u_2)\otimes \ldots T(u_k)$
If $\displaystyle \sigma$ is a permutation, then $\displaystyle (u_1\otimes \ldots u_k)^\sigma=u_{\sigma^{-1}(1)}\otimes \ldots \otimes u_{\otimes k}$ and thus we can immediately verify the identity
$\displaystyle T(u_1\otimes \ldots u_k)^\sigma=T((u_1\otimes \ldots u_k)^\sigma)$
because both sides will equal $\displaystyle T(u_{\sigma^{-1}(1)})\otimes \ldots T(u_{\sigma^{-1}(k)})$
Because pure tensors span the k-tensor power space, we conclude that
$\displaystyle T(v^\sigma)=T(v)^\sigma$ (1)
Now let's get back to the problem.
By definition, $\displaystyle T^*(S^\sigma)(x)=S^\sigma(T(x))$ which is $\displaystyle S(T(x))^\sigma$
Now by two consecutive applications of (1), $\displaystyle S(T(x))^\sigma=S(T(x)^\sigma)=S(T(x^\sigma))$ and that is, by definition, $\displaystyle (T^*S)(x^\sigma)$ - which again by definition equals $\displaystyle (T^*S)^\sigma(x)$