I've seen in the literature the notation $C$ with some additional specifications for the contraction maps of all sorts, but the amount of decorations on the symbol $C$ varied depending on the context. See, e.g., A.Gray, Tubes, p.56, where these maps are used in the case of somewhat special tensors, and therefore the notation is simpler.
In general, there is a whole family of uniquely defined maps
$$
C^{(r,s)}_{p,q} \colon \otimes^{r}_{s} V \to \otimes^{r-1}_{s-1} V
$$
which are collectively called tensor contractions ($1 \le p \le r, 1 \le q \le s$).
These maps are uniquely characterized by making the following diagrams commutative:
$$
\require{AMScd}
\begin{CD}
\times^{r}_{s} V @> {P^{(r,s)}_{p,q}} >> \times^{r-1}_{s-1} V\\
@V{\otimes^{r}_{s}}VV @VV{\otimes^{r-1}_{s-1}}V \\
\otimes^{r}_{s} V @>{C^{(r,s)}_{p,q}}>> \otimes^{r-1}_{s-1} V
\end{CD}
$$
Explanations are in order.
Recall that the tensor products $\otimes^{r}_{s} V$ are equipped with the universal maps
$$
\otimes^{r}_{s} \colon \times^{r}_{s} V \to \otimes^{r}_{s} V
$$
where $\times^{r}_{s} V := ( \times^r V) \times (\times^s V^*)$.
Besides that, there is a canonical pairing $P$ between a vector space $V$ and its dual:
$$
P \colon V \times V^* \to \mathbb{R} \colon (v, \omega) \mapsto \omega(v)
$$
Notice that map $P$ is bilinear and can be extended to a family of multilinear maps
$$
P^{(r,s)}_{p,q} \colon \times^{r}_{s} V \to \times^{r-1}_{s-1} V
$$
by the formula:
$$
P^{(r,s)}_{p,q} (v_1, \dots, v_p, \dots, v_r, \omega_1, \dots, \omega_q, \dots, \omega_s) = \omega_q (v_p) (v_1, \dots, \widehat{v_p}, \dots, v_r, \omega_1, \dots, \widehat{\omega_q}, \dots, \omega_s)
$$
where a hat means omission.
Since maps $P^{(r,s)}_{p,q}$ are multilinear, the universal property of the maps $\otimes^{r}_{s}$ implies that there are uniquely defined maps
$$
\tilde{P}^{(r,s)}_{p,q} \colon \otimes^{r}_{s} V \to \times^{r-1}_{s-1} V
$$
and then the maps $C^{(r,s)}_{p,q}$ are given by
$$
C^{(r,s)}_{p,q} := \otimes^{r-1}_{s-1} \circ \tilde{P}^{(r,s)}_{p,q}
$$
Best Answer
I came up with this.
Let $X$ be $I_1\times\cdots \times I_N$ tensor and $A^{(n)}$ $J_n\times I_n$ matrices. By matricization definition, $y_{j_1\cdots j_N}$ element of $J_1\times\cdots\times J_N$ tensor $Y$ maps to $y_{j_nk}$ element of matrix $Y_{(n)}$, with $$ k=1+\sum_{\substack{l=1\\l\neq n}}^N (j_l-1)\prod_{\substack{m=1\\m\neq n}}^{l-1} J_m. $$ We will prove the statement by showing that every element $$ \left( X \times_1 A^{(1)} \times_2 A^{(2)}\times_3 \cdots \times_N A^{(N)}\right)_{j_1\cdots j_N} $$ is the same as element $$ \left[{A}^{(n)}{X}_{(n)}\left( {A}^{(N)} \otimes \cdots \otimes {A}^{(n+1)} \otimes {A}^{(n-1)} \otimes \cdots \otimes {A}^{(1)} \right)^T\right]_{j_nk}, $$ with $k$ as stated.
From $n$-mode product definition, we have $$ \left( {X} \times_1 {A}^{(1)} \times_2 {A}^{(2)}\times_3 \cdots \times_N {A}^{(N)}\right)_{j_1\cdots j_N}=\sum_{i_1=1}^{I_1}\cdots\sum_{i_N=1}^{I_N} x_{i_1\cdots i_N} a^{(1)}_{j_1i_1}\cdots a^{(N)}_{j_Ni_N}. $$ On the other hand, by denoting ${M}_n=\left( {A}^{(N)} \otimes \cdots \otimes {A}^{(n+1)} \otimes {A}^{(n-1)} \otimes \cdots \otimes {A}^{(1)} \right)^T$, we have \begin{align} \left({A}^{(n)}{X}_{(n)}{M}_n\right)_{j_nk}&={A}^{(n)}[\,j_n\,,\,:\,]\left({X}_{(n)}{M}_n\right)[\,:\,,\,k\,] \nonumber \\ &=\sum_{i_n=1}^{I_n} a^{(n)}_{j_ni_n}\left({X}_{(n)}{M}_n\right)[\,i_n\,,\,k\,] \nonumber \\ &=\sum_{i_n=1}^{I_n} a^{(n)}_{j_ni_n}\sum_{i=1}^{\hat{I}_n}{X}_{(n)}[\,i_n\,,\,i\,]{M}_n[\,i\,,\,k\,], \label{eq:proofjk} \end{align} with $\hat{I}_n=I_1\cdots I_{n-1}I_{n+1}\cdots I_N$. Now, ${X}_{(n)}[\,i_n\,,\,i\,]=x_{i_1\cdots i_N}$, with $$ i=1+\sum_{\substack{l=1\\l\neq n}}^N (i_l-1)\prod_{\substack{m=1\\m\neq n}}^{l-1} I_m. $$ The same $i$ stands in ${M}_n[\,i\,,\,k\,]$, which we easily get from the definition of Kronecker product $$ {M}_n[\,i\,,\,k\,]=\tilde{a}^{(N)}_{i_Nj_N}\cdots \tilde{a}^{(n+1)}_{i_{n+1}j_{n+1}}\tilde{a}^{(n+1)}_{i_{n-1}j_{n-1}}\cdots\tilde{a}^{(1)}_{i_1j_1}, $$ with $\tilde{a}^{(m)}_{i_mj_m}$ denoting element of ${{A}^{(m)}}^T$. Using these conclusions, we can rewrite the above as $$ \left({A}^{(n)}{X}_{(n)}{M}_n\right)_{j_nk}=\sum_{i_1=1}^{I_1}\cdots\sum_{i_N=1}^{I_N} x_{i_1\cdots i_N} a^{(1)}_{j_1i_1}\cdots a^{(N)}_{j_Ni_N}, $$ which completes the proof.