= Solution
Use the ordered alphabet
$$
\mathcal A=(1,2,\ldots,n,\bar n,\overline{n-1},\ldots,\bar1),
$$
with <weights> $\operatorname{wt}(i)=\varepsilon_i$ and $\operatorname{wt}(\bar i)=-\varepsilon_i$. The <crystal of the defining symplectic representation> is the chain
$$
1\xrightarrow{1}2\xrightarrow{2}\cdots\xrightarrow{n-1}n\xrightarrow{n}\bar n\xrightarrow{n-1}\cdots\xrightarrow{2}\bar2\xrightarrow{1}\bar1.
$$
The label of an arrow is its <simple root> color; a <Kashiwara operator> $\widetilde f_i$ follows an arrow of color $i$ and lowers the <weight> by $\alpha_i$.
Here is a grid description of the entire <tensor product of crystals>, valid for every $n$. Put one vertex $a\otimes b$ at row $a$ and column $b$, in the alphabet order above. Let $P_i$ be the set of chain positions with outgoing color $i$, and $Q_i$ the positions with incoming color $i$:
$$
P_i=\{i,2n-i\},\quad Q_i=\{i+1,2n-i+1\}\quad(i<n),\qquad P_n=\{n\},\quad Q_n=\{n+1\}.
$$
Their indicators are $\varphi_i$ and $\varepsilon_i$. Use the <crystal tensor-product rule>
$$
\widetilde f_i(a\otimes b)=\begin{cases}
\widetilde f_i(a)\otimes b,&\varphi_i(a)>\varepsilon_i(b),\\
a\otimes\widetilde f_i(b),&\varphi_i(a)\le\varepsilon_i(b).
\end{cases}
$$
An arrow is omitted if the selected factor has no outgoing arrow of that color. Thus the complete grid has a downward color-$i$ arrow when $a\in P_i$ and $b\notin Q_i$, and a rightward color-$i$ arrow when $b\in P_i$ and $a\notin P_i$. There are no other arrows. These two coordinate rules specify every vertex and every edge of the requested general-rank drawing. The following figure displays the full grid at $n=3$, including all components, together with the standard chain; increasing $n$ uses exactly the same grid construction.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/iii/paper-6-crystals.png]
{title=Defining C3 crystal and its complete tensor-square crystal with all colored arrows and the three highest-weight components}
For completeness, highest vertices can be read from the raising rule, which selects the first factor when $\varphi_i(a)\ge\varepsilon_i(b)$. If $a$ has an incoming edge of color $i$, then $\varphi_i(a)=0$; either the raising operator acts nontrivially on $a$, or $\varepsilon_i(b)=1$ and it acts nontrivially on $b$. Thus a highest vertex must have $a=1$. At $a=1$, all $\varphi_i(a)$ vanish except $\varphi_1(a)=1$. To be highest, the incoming color of $b$, if any, must consequently be $1$. For $n\ge2$, the possibilities are exactly
$$
1\otimes1,\qquad 1\otimes2,\qquad 1\otimes\bar1.
$$
Their <weights> are $2\varepsilon_1=2\omega_1$, $\varepsilon_1+\varepsilon_2=\omega_2$, and zero. Each occurs once. Connected highest-weight components of the <crystal basis> describe the <irreducible representations> in the <tensor product of Lie algebra representations>, so
$$
\boxed{V\otimes V\cong L(2\omega_1)\oplus L(\omega_2)\oplus L(0)\qquad(n\ge2).}
$$
This is the <tensor-square decomposition of the defining symplectic representation>. The <symmetric square> is $L(2\omega_1)$; the <exterior square> is the <direct sum> of its invariant inverse-form line and the <primitive exterior square> $L(\omega_2)$. Their <dimensions> are respectively $n(2n+1)$, $n(2n-1)-1$, and one, summing to $4n^2=\dim(V\otimes V)$. At $n=1$, the chain has two vertices and the highest tensor vertices are only $1\otimes1$ and $1\otimes\bar1$: the decomposition is $L(2\omega_1)\oplus L(0)$, of <dimensions> three and one.
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