Let $p\in\mathbb N$ and $n_1,\ldots,n_p\in\mathbb N$. Assume the tensor product space $\bigotimes_{i=1}^p\mathbb R^{n_i}$ is equipped with the unique inner product $\langle\;\cdot\;,\;\cdot\;\rangle_{\bigotimes_{i=1}^p\mathbb R^{n_i}}$ satisfying $$\left\langle\bigotimes_{i=1}^px^{(i)},\bigotimes_{i=1}^py^{(i)}\right\rangle_{\bigotimes_{i=1}^p\mathbb R^{n_i}}=\prod_{i=1}^p\left\langle x^{(i)},y^{(i)}\right\rangle\;\;\;\text{for all }x^{(i)}\in\mathbb R^{n_i},\tag1$$ where $\langle\;\cdot\;,\;\cdot\;\rangle$ denotes the Euclidean inner product. For any $I\subseteq\{1,\ldots,p\}$ there is a canonical$^1$ isomorphism $$M_I:\bigotimes_{i=1}^p\mathbb R^{n_i}\to\mathbb R^{\prod_{i\in I}n_i\times\prod_{i\in I^c}n_i}$$
The question is: Are the isomorphisms $M_I$ isometric? It clearly depends on the matrix norm chosen on $\mathbb R^{\prod_{i\in I}n_i\times\prod_{i\in I^c}n_i}$. I would like to show the isometricness for the Frobenius norm$^2$.
$^1$ Assume that we have fixed an isomorphism converting a multiindex into a single index.
$^2$ if $\left(e^{(n)}_1,e^{(n)}_n\right)$ denotes the standard basis of $\mathbb R^n$ and $A,B\in\mathbb R^{m\times n}$, then the Frobenius inner product of $A,B$ is given by $$\langle A,B\rangle_F=\sum_{i=1}^n\left\langle Ae^{(n)}_i,Be^{(n)}_i\right\rangle.\tag2$$