Solution:
The equation of a common tangent line to the graphs of $f(x)$ and $g(x)$ at points $(x_{1}, f(x_{1}))$ and $(x_{2}, g(x_{2}))$ has the form
$$y = f(x_{1}) + f'(x_{1})(x - x_{1}) = g(x_{2}) + g'(x_{2})(x - x_{2})$$
Hence $f'(x_{1}) = g'(x_{2})$ and $f(x_{1}) - f'(x_{1}) x_{1} = g(x_{2}) - g'(x_{2}) x_{2}$. Since $f'(x) = 2x - 2a$ and $g'(x) = -2x$, we get that $x_{1} + x_{2} = a$ and $x_{1}^{2} + x_{2}^{2} = 1$. Then $x_{1} x_{2} = \frac{a^{2} - 1}{2}$ and hence $x_{1}$ and $x_{2}$ are the roots of the quadratic equation $x^{2} - a x + \frac{a^{2} - 1}{2} = 0$. Since the graphs of $f(x) = x^{2} - 2a x$ and $g(x) = -x^{2} - 1$ have two common tangent lines, it follows that they are disjoint, $x_{1} \neq x_{2}$, $a^{2} < 2$, and the tangent points are $M(x_{1}, f(x_{1})), N(x_{2}, f(x_{2})), P(x_{2}, g(x_{2}))$ and $Q(x_{1}, g(x_{1}))$. Then
$$\begin{aligned} PQ^{2} & = (x_{1} - x_{2})^{2} + (g(x_{1}) - g(x_{2}))^{2} \\ & = (x_{1}^{2} + x_{2}^{2} - 2 x_{1} x_{2})^{2} (1 + x_{1} + x_{2}) = (2 - a^{2})(1 + a^{2}) \\ MQ & = |f(x_{1}) - g(x_{1})| = |x_{1}^{2} - 2a x_{1} + x_{1}^{2} + 1| = 2 - a^{2} \end{aligned}$$
and similarly
$$MN^{2} = (2 - a^{2})(1 + a^{2}), \quad NP = 2 - a^{2}$$
Hence $MNPQ$ is a parallelogram with perimeter
$$2\left(\sqrt{(2 - a^{2})(1 + a^{2})} + 2 - a^{2}\right) = 6$$
Then $\sqrt{(2 - a^{2})(1 + a^{2})} = 1 + a^{2}$ and we get that $a = \pm \frac{\sqrt{2}}{2}$.