The vertices $V$ of a centrally symmetric hexagon in the complex plane are given by
$$V=\left\{   \sqrt{2}i,-\sqrt{2}i, \frac{1}{\sqrt{8}}(1+i),\frac{1}{\sqrt{8}}(-1+i),\frac{1}{\sqrt{8}}(1-i),\frac{1}{\sqrt{8}}(-1-i) \right\}.$$
For each $j$, $1\leq j\leq 12$, an element $z_j$ is chosen from $V$ at random, independently of the other choices. Let $P={\prod}_{j=1}^{12}z_j$ be the product of the $12$ numbers selected. What is the probability that $P=-1$?

$\textbf{(A) } \dfrac{5\cdot11}{3^{10}} \qquad \textbf{(B) } \dfrac{5^2\cdot11}{2\cdot3^{10}} \qquad \textbf{(C) } \dfrac{5\cdot11}{3^{9}} \qquad \textbf{(D) } \dfrac{5\cdot7\cdot11}{2\cdot3^{10}} \qquad \textbf{(E) } \dfrac{2^2\cdot5\cdot11}{3^{10}}$
1. **Identify the vertices and their properties:**
   The vertices of the hexagon are given by:
   \[
   V = \left\{ \sqrt{2}i, -\sqrt{2}i, \frac{1}{\sqrt{8}}(1+i), \frac{1}{\sqrt{8}}(-1+i), \frac{1}{\sqrt{8}}(1-i), \frac{1}{\sqrt{8}}(-1-i) \right\}
   \]
   The magnitudes of these vertices are:
   \[
   |\sqrt{2}i| = \sqrt{2}, \quad |-\sqrt{2}i| = \sqrt{2}, \quad \left|\frac{1}{\sqrt{8}}(1+i)\right| = \frac{1}{2}, \quad \left|\frac{1}{\sqrt{8}}(-1+i)\right| = \frac{1}{2}, \quad \left|\frac{1}{\sqrt{8}}(1-i)\right| = \frac{1}{2}, \quad \left|\frac{1}{\sqrt{8}}(-1-i)\right| = \frac{1}{2}
   \]

2. **Determine the condition for the magnitudes:**
   When multiplying 12 complex numbers, their magnitudes multiply. Suppose we have \( n \) of the numbers \( \pm \sqrt{2}i \); then we must have:
   \[
   \left(\sqrt{2}\right)^n \cdot \left(\frac{1}{2}\right)^{12-n} = 1
   \]
   Solving for \( n \):
   \[
   2^{n/2} \cdot 2^{-(12-n)} = 1 \implies 2^{n/2 - 12 + n} = 1 \implies 2^{3n/2 - 12} = 1 \implies 3n/2 - 12 = 0 \implies n = 8
   \]
   Thus, we need exactly 8 of the numbers \( \pm \sqrt{2}i \) and 4 of the numbers \( \frac{1}{\sqrt{8}}(\pm 1 \pm i) \).

3. **Determine the condition for the angles:**
   We require:
   \[
   \sum \theta \equiv \pi \mod 2\pi
   \]
   Let \( z = e^{i\pi/4} \). The angles we have available are \( \{z^1, z^2, z^3, z^5, z^6, z^7\} \). We must choose exactly 8 angles from the set \( \{z^2, z^6\} \) and exactly 4 from the set \( \{z^1, z^3, z^5, z^7\} \).

4. **Count the number of good combinations:**
   If we found a good combination where we had \( a_i \) of each angle \( z^i \), then the amount this would contribute to our count would be:
   \[
   \binom{12}{4,8} \cdot \binom{8}{a_2, a_6} \cdot \binom{4}{a_1, a_3, a_5, a_7}
   \]
   We want to add these all up. We proceed by generating functions.

5. **Generating function approach:**
   Consider:
   \[
   (t_2x^2 + t_6x^6)^8 (t_1x^1 + t_3x^3 + t_5x^5 + t_7x^7)^4
   \]
   The expansion will be of the form:
   \[
   \sum_i \left( \sum_{\sum a = i} \binom{8}{a_2, a_6} \binom{4}{a_1, a_3, a_5, a_7} t_1^{a_1} t_2^{a_2} t_3^{a_3} t_5^{a_5} t_6^{a_6} t_7^{a_7} x^i \right)
   \]
   Note that if we reduced the powers of \( x \) mod \( 8 \) and fished out the coefficient of \( x^4 \) and plugged in \( t_i = 1 \ \forall \, i \) (and then multiplied by \( \binom{12}{4,8} \)), then we would be done. Since plugging in \( t_i = 1 \) doesn't affect the \( x \)'s, we do that right away. The expression then becomes:
   \[
   x^{20}(1 + x^4)^8 (1 + x^2 + x^4 + x^6)^4 = x^{20}(1 + x^4)^{12}(1 + x^2)^4 = x^4(1 + x^4)^{12}(1 + x^2)^4
   \]
   where the last equality is true because we are taking the powers of \( x \) mod \( 8 \).

6. **Roots of unity filter:**
   Let \( [x^n]f(x) \) denote the coefficient of \( x^n \) in \( f(x) \). Note:
   \[
   [x^4] x^4(1 + x^4)^{12}(1 + x^2)^4 = [x^0](1 + x^4)^{12}(1 + x^2)^4
   \]
   We use the roots of unity filter, which states:
   \[
   \text{terms of } f(x) \text{ that have exponent congruent to } k \text{ mod } n = \frac{1}{n} \sum_{m=1}^n \frac{f(z^m x)}{z^{mk}}
   \]
   where \( z = e^{i\pi/n} \). In our case \( k = 0 \), so we only need to find the average of the \( f(z^m x) \)'s.

7. **Evaluate the roots of unity filter:**
   \[
   \begin{align*}
   z^0 &\Longrightarrow (1 + x^4)^{12}(1 + x^2)^4, \\
   z^1 &\Longrightarrow (1 - x^4)^{12}(1 + ix^2)^4, \\
   z^2 &\Longrightarrow (1 + x^4)^{12}(1 - x^2)^4, \\
   z^3 &\Longrightarrow (1 - x^4)^{12}(1 - ix^2)^4, \\
   z^4 &\Longrightarrow (1 + x^4)^{12}(1 + x^2)^4, \\
   z^5 &\Longrightarrow (1 - x^4)^{12}(1 + ix^2)^4, \\
   z^6 &\Longrightarrow (1 + x^4)^{12}(1 - x^2)^4, \\
   z^7 &\Longrightarrow (1 - x^4)^{12}(1 - ix^2)^4.
   \end{align*}
   \]
   We plug in \( x = 1 \) and take the average to find the sum of all coefficients of \( x^{\text{multiple of 8}} \). Plugging in \( x = 1 \) makes all of the above zero except for \( z^0 \) and \( z^4 \). Averaging, we get \( 2^{14} \).

8. **Calculate the final probability:**
   The total number of ways to choose 12 complex numbers is \( 6^{12} \). The number of good combinations is:
   \[
   \frac{\binom{12}{4,8}}{6^{12}} \cdot 2^{14}
   \]
   Simplifying:
   \[
   \binom{12}{4,8} = \frac{12!}{4!8!} = \frac{12 \cdot 11 \cdot 10 \cdot 9}{4 \cdot 3 \cdot 2 \cdot 1} = 495
   \]
   Thus, the probability is:
   \[
   \frac{495 \cdot 2^{14}}{6^{12}} = \frac{495 \cdot 2^{14}}{2^{12} \cdot 3^{12}} = \frac{495 \cdot 2^2}{3^{12}} = \frac{495 \cdot 4}{3^{12}} = \frac{1980}{3^{12}} = \frac{2^2 \cdot 5 \cdot 11}{3^{10}}
   \]

The final answer is \(\boxed{\frac{2^2 \cdot 5 \cdot 11}{3^{10}}}\)