Consider two arbitrary points $X$ and $Y$ and let $x$ and $y$ be the corresponding numbers. Choose points $I$ and $J$ on the line $XY$ such that $XY = YI = IJ$. Let $X'$ on the line $XY$ be such that $XI = JX'$. Consider the plane $\lambda$ perpendicular to $IJ$ and passing through the midpoint of $IJ$. Let $ABCX$ and $ABCX'$ be two equal regular triangular pyramids with base $ABC$ in the plane $\lambda$ having incenters $I$ and $J$.

Since $n_A n_B n_C n_X = n_I$ and $n_A n_B n_C n_{X'} = n_J$ we have that
$$n_X = \frac{n_{X'} \cdot n_I}{n_J}.$$
Move the plane $\lambda$ towards point $I$ and consider the spheres $S_I$ and $S_J$ with centers $I$ and $J$ respectively that are tangent to $\lambda$. Let $A_1B_1C_1X'$ be a regular triangular pyramid with base $A_1B_1C_1$ in $\lambda$ and insphere $S_J$. The regular triangular pyramid with base $A_1B_1C_1$ and insphere $S_I$ has vertex $X_1$. It follows from the above that
$$n_{X_1} = \frac{n_{X'} \cdot n_I}{n_J} = n_X.$$
When $\lambda$ moves towards $I$ the radius of the sphere $S_I$ tends to zero and $\triangle A_1B_1C_1$ tends to a triangle which is the base of a regular triangular pyramid with vertex $X'$ and inscribed sphere with center $J$ and radius $IJ$.

We conclude that $X_1$ tends to $I$. Continuity arguments show that all inner points on the segment $XI$ (with point $X$) are assigned with the same number.

Thus, $x = y$ and all numbers are equal. It follows from $x^4 = x$ and $x \neq 0$ that $x = 1$.