Let $ABCDEFG$ be a regular heptagon with center $O$.  Let $M$ be the centroid of triangle $ABD$.  Find $\cos^2 \angle GOM$.
Let $\omega = e^{2 \pi i/7}$.  Then $\omega^7 = 1$, so $\omega^7 - 1 = 0$, which factors as
\[(\omega - 1)(\omega^6 + \omega^5 + \omega^4 + \omega^3 + \omega^2 + \omega + 1) = 0.\]Since $\omega \neq 1$, $\omega$ satisfies
\[\omega^6 + \omega^5 + \omega^4 + \omega^3 + \omega^2 + \omega + 1 = 0.\]We place heptagon $ABCDEFG$ in the plane, so that $G$ is at 1, $A$ is at $\omega$, $B$ is at $\omega^2$, and so on.

[asy]
unitsize(2 cm);

pair A, B, C, D, E, F, G, M, O;

G = dir(0);
A = dir(360/7);
B = dir(2*360/7);
C = dir(3*360/7);
D = dir(4*360/7);
E = dir(5*360/7);
F = dir(6*360/7);
M = (A + B + D)/3;

draw(A--B--C--D--E--F--G--cycle);
draw(B--D--A);
draw(M--O--G--cycle);

label("$1$", G, G);
label("$\omega$", A, A);
label("$\omega^2$", B, B);
label("$\omega^3$", C, C);
label("$\omega^4$", D, D);
label("$\omega^5$", E, E);
label("$\omega^6$", F, F);
dot("$m$", M, N);
dot("$0$", (0,0), SW);
[/asy]

Then the centroid of triangle $ABD$ is at
\[m = \frac{\omega + \omega^2 + \omega^4}{3}.\]Now, by the law of cosines,
\[\cos \angle GOM = \frac{OG^2 + OM^2 - GM^2}{2 \cdot OG \cdot OM}.\]We see that $OG = 1$, and
\begin{align*}
OM^2 &= |m|^2 \\
&= m \overline{m} \\
&= \frac{\omega + \omega^2 + \omega^4}{3} \cdot \frac{1/\omega + 1/\omega^2 + 1/\omega^4}{3} \\
&= \frac{(\omega + \omega^2 + \omega^4)(\omega^6 + \omega^5 + \omega^3)}{9} \\
&= \frac{\omega^7 + \omega^6 + \omega^4 + \omega^8 + \omega^7 + \omega^5 + \omega^{10} + \omega^9 + \omega^7}{9} \\
&= \frac{1 + \omega^6 + \omega^4 + \omega + 1 + \omega^5 + \omega^3 + \omega^2 + 1}{9} \\
&= \frac{\omega^6 + \omega^5 + \omega^4 + \omega^3 + \omega^2 + \omega + 3}{9} \\
&= \frac{2}{9}.
\end{align*}Also,
\begin{align*}
GM^2 &= |1 - m|^2 \\
&= (1 - m)(1 - \overline{m}) \\
&= 1 - m - \overline{m} + m \overline{m} \\
&= 1 - \frac{\omega + \omega^2 + \omega^4}{3} - \frac{\omega^6 + \omega^5 + \omega^3}{3} + \frac{2}{9} \\
&= \frac{11}{9} - \frac{\omega^6 + \omega^5 + \omega^4 + \omega^3 + \omega^2 + \omega}{3} \\
&= \frac{11}{9} + \frac{1}{3} \\
&= \frac{14}{9}.
\end{align*}Then $OM = \sqrt{2}/3$, so
\begin{align*}
\cos \angle GOM &= \frac{OG^2 + OM^2 - GM^2}{2 \cdot OG \cdot OM} \\
&= \frac{1 + 2/9 - 14/9}{2 \cdot 1 \cdot \sqrt{2}/3} \\
&= \frac{-3/9}{2 \sqrt{2}/3} \\
&= -\frac{1}{2 \sqrt{2}},
\end{align*}which means
\[\cos^2 \angle GOM = \left( -\frac{1}{2 \sqrt{2}} \right)^2 = \boxed{\frac{1}{8}}.\]