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Okay, so I think this is what's going on. The elements of $V^{\otimes n}$ are naturally identified with multilinear functions on $V^{\ast}$. Among these functions are the alternating multilinear functions on $V^{\ast}$, which are naturally identified with the elements of $\Lambda^n(V^{\ast})^{\ast}$, so this gives a natural inclusion $\Lambda^n(V^{\ast})^{\ast} \to V^{\otimes n}$. The situation is similar for symmetric tensors. I think this inclusion is what I'm looking for.

One can compose the above inclusion with the quotient $V^{\otimes n} \to \Lambda^n(V)$, and this gives a natural map $\Lambda^n(V^{\ast})^{\ast} \to \Lambda^n(V)$. In characteristic zero this map is an isomorphism, but in positive characteristic there are problems, and in any case it behaves in a slightly unexpected way with respect to a basis of $V$ (because of the issues Greg Kuperberg brought up in the linked MO question).

Here is what happens in the simplest nontrivial case. Let $V$ be two-dimensional with basis $e_1, e_2$. Then $V^{\otimes 2}$ inherits a natural basis $e_i \otimes e_j, 1 \le i, j \le 2$ and the image of this basis gives a basis $e_1 \wedge e_2$ of $\Lambda^2 V$. The dual $V^{\ast}$ inherits a dual basis $e_1^{\ast}, e_2^{\ast}$ giving a basis $e_1^{\ast} \wedge e_2^{\ast}$ of $\Lambda^2 V$, and dualizing one more time gives a dual basis $(e_1^{\ast} \wedge e_2^{\ast})^{\ast}$. The natural inclusion above sends $(e_1^{\ast} \wedge e_2^{\ast})^{\ast}$ to $e_1 \otimes e_2 - e_2 \otimes e_1$.

But now the natural map $\Lambda^2(V^{\ast})^{\ast} \to \Lambda^2(V)$ sends $(e_1^{\ast} \wedge e_2^{\ast})^{\ast}$ to $2 e_1 \wedge e_2$. So natural bases do not behave in the expected way with respect to these constructions, and one must insert some factorials...

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Okay, so I think this is what's going on. The elements of $V^{\otimes n}$ are naturally identified with multilinear functions on $V^{\ast}$. Among these functions are the alternating multilinear functions on $V^{\ast}$, which are naturally identified with the elements of $\Lambda^n(V^{\ast})^{\ast}$, so this gives a natural inclusion $\Lambda^n(V^{\ast})^{\ast} \to V^{\otimes n}$. The situation is similar for symmetric tensors. I think this inclusion is what I'm looking for.

One can compose the above inclusion with the quotient $V^{\otimes n} \to \Lambda^n(V)$, and this gives a natural map $\Lambda^n(V^{\ast})^{\ast} \to \Lambda^n(V)$. In characteristic zero this map is an isomorphism, but in positive characteristic there are problems, and in any case it behaves in a slightly unexpected way with respect to a basis of $V$ (because of the issues Greg Kuperberg brought up in the linked MO question).

Here is what happens in the simplest nontrivial case. Let $V$ be two-dimensional with basis $e_1, e_2$. Then $V^{\otimes 2}$ inherits a natural basis $e_i \otimes e_j, 1 \le i, j \le 2$ and the image of this basis gives a basis $e_1 \wedge e_2$ of $\Lambda^2 V$. The dual $V^{\ast}$ inherits a dual basis $e_1^{\ast}, e_2^{\ast}$ giving a basis $e_1^{\ast} \wedge e_2^{\ast}$ of $\Lambda^2 V$, and dualizing one more time gives a dual basis $(e_1^{\ast} \wedge e_2^{\ast})^{\ast}$.

But now the natural map $\Lambda^2(V^{\ast})^{\ast} \to \Lambda^2(V)$ sends $(e_1^{\ast} \wedge e_2^{\ast})^{\ast}$ to $2 e_1 \wedge e_2$. So natural bases do not behave in the expected way with respect to these constructions, and one must insert some factorials...