Note that both maps below satisfy the universal property of the tensor product.
\begin{align*}
\mathbb{R}^2 \times \mathbb{R}^2 &\rightarrow \mathbb{R}^{2 \times 2} &\mathbb{R}^2 \times \mathbb{R}^2 &\rightarrow \mathbb{R}^{2 \times 2} \\
\begin{bmatrix}
a_1 \\
a_2 \\
\end{bmatrix} \times
\begin{bmatrix}
b_ 1 \\
b_2 \\
\end{bmatrix}
&\mapsto
\begin{bmatrix}
a_1 b_ 1 & a_1 b_2 \\
a_2 b_1 & a_2 b_2 \\
\end{bmatrix}
&\begin{bmatrix}
a_1 \\
a_2 \\
\end{bmatrix} \times
\begin{bmatrix}
b_ 1 \\
b_2 \\
\end{bmatrix}
&\mapsto
\begin{bmatrix}
a_1 b_ 2 & a_1 b_1 \\
a_2 b_2 & a_2 b_1 \\
\end{bmatrix} \\
(a_i) \times (b_i) &\mapsto (c_{ij} = a_i b_j) &(a_i) \times (b_i) &\mapsto (c_{ij} = a_i b_{\tau(j)}) \text{ where $\tau = (12) \in S_2$}
\end{align*}
Similarly, for $(\sigma_1, \sigma_2, \dots, \sigma_d) \in S_{n_1} \times S_{n_2} \times \dots S_{n_d}$, the map
\begin{gather*}
K^{n_1} \times K^{n_2} \times \dots \times K^{n_d} \ \xrightarrow{\Gamma_{(\sigma_1, \sigma_2, \dots, \sigma_d)}} \ K^{n_1 \times n_2 \times \dots \times n_d} \\
(x_i^1) \times (x_i^2) \times \dots \times (x_i^d) \ \mapsto \ (c_{{i_1}{i_2}\dots{i_d}} = x^1_{\sigma_1(i_1)} x^2_{\sigma_2(i_2)} \dots x^d_{\sigma_d(i_d)})
\end{gather*} satisfies the universal property.
Are these the only maps that satisfy the universal property? Formally, is it true that:
if $K^{n_1} \times K^{n_2} \times \dots \times K^{n_d} \ \xrightarrow{\Gamma} \ K^{n_1 \times n_2 \times \dots \times n_d}$ satisfies the universal property of the tensor product, then $\Gamma = \Gamma_{(\sigma_1, \sigma_2, \dots, \sigma_d)}$ for some $(\sigma_1, \sigma_2, \dots, \sigma_d) \in S_{n_1} \times S_{n_2} \times \dots S_{n_d}$.
I asked a similar question here Proof of unique coordinatization of tensor space once bases are chosen but I think I confused people in how the question was stated, so I am writing it in a different way here. Thank you for any help you can offer.
No, there are many more such "tensor maps". For instance, (I think, as you didn't specify exactly which tensor product property you wanted) you can compose with any automorphism of $\mathbb{R}^{2 \times 2}$ to create a different tensor map.
For instance, for any invertible matrix $P$, $(a,b) \longmapsto P^{-1}\Gamma_{\operatorname{id},\operatorname{id}}(a,b)P$ should satisfy the property.
For a concrete example, take $P=\begin{bmatrix}0 & 1\\-1 & 0\end{bmatrix}$ and consider the map $$\begin{bmatrix}a_1\\ a_2\end{bmatrix} \times \begin{bmatrix}b_1\\ b_2\end{bmatrix} \longmapsto \begin{bmatrix} a_2b_2 & -a_2b_1 \\-a_1b_2 & a_1b_1\end{bmatrix}.$$