Assume that there are $n+1$ vectors of dimension $n$ and all angles of any two vectors equal, is the cosine of the angle $-\frac{1}{n}$? When $n=2$, the $3$ vectors construct a regular triangle, and the consine of the angle is $-\frac{1}{2}$
When $n=3$, the $4$ vectors construct a regular tetrahedon, and the consine of the angle is $-\frac{1}{3}$
Is it true for any $n$? I guess so, but how to prove?
 A: You can do this without needing any specific assignments of coordinates except for having one node at the origin. This is how you do it.
Let one node be at the origin (obviously). Now, you have a set of $n$ vectors pointing to each of the other nodes. We call these $\mathbf{v}_i$. As all distances between nodes are of the same length, we have that $|\mathbf{v}_i|=|\mathbf{v}_j-\mathbf{v}_i|=C$ for $j\neq i$. Let the centre be $\mathbf{c}=\frac1{n+1}\sum_{i=1}^n \mathbf{v}_i$.
Now, we consider the centre, the origin node, and one other node - without loss of generality, we use $\mathbf{v}_1$. These three points form a triangle, and from the Law of Cosines, we have
$$
\mathbf{c}\cdot(\mathbf{c}-\mathbf{v}_1) = |\mathbf{c}||\mathbf{c}-\mathbf{v}_1|\cos \theta
$$
Now, noting the symmetry of the system, we have
$$
\mathbf{c}\cdot\mathbf{v}_1 = \frac{C^2+(n-1)\mathbf{v_2}\cdot\mathbf{v_1}}{n+1}
$$
Each face forms an equilateral triangle, so $\mathbf{v}_2\cdot\mathbf{v}_1=\frac{C^2}2$. This allows us to write
$$
\mathbf{c}\cdot\mathbf{v}_1 = \frac{C^2}2
$$
Similarly, we have
$$
|\mathbf{c}|^2 = n\frac{C^2+(n-1)\mathbf{v}_2\cdot\mathbf{v}_1}{(n+1)^2}=\frac{nC^2}{2(n+1)}
$$
Substituting these into our equation, and noting that $|\mathbf{c}-\mathbf{v}_1|=|\mathbf{c}|=\frac{C}{\sqrt{2}}$, we have
$$
\cos\theta = \frac{\frac{nC^2}{2(n+1)}-\frac{C^2}2}{\frac{nC^2}{2(n+1)}}=\frac{\frac{n}{n+1}-1}{\frac{n}{n+1}} = \frac{\ \frac{-1}{n+1}\ }{\frac{n}{n+1}} = -\frac1n
$$
A: Regular simplex
If you have $n+1$ vectors in $n$ dimensions, and all angles between any two vectors are equal, then they form a regular simplex. For $n=2$ that's a regular triangle, for $n=3$ it's a regular tetrahedron, and so on. You are asking for that common angle, i.e. the angle between two of the lines formed by the center of the simplex and its corners. In your description the center is the origin of the coordinate system, but that choice is not essential for your question: you might as well choose a different coordinate system as long as you keep looking at those lines through the center of the simplex.
Choosing a different coordinate system is particularly useful if you increase the dimension by one: in $n+1$ dimensions there are $n+1$ unit vectors. They all lie in a single hyperplane, i.e. in a $n$-dimensional affine space, but the coordinates in $n+1$ dimensions are a lot more symmetric and therefore a lot easier to handle. The Wikipedia article calls this choice of coordinates the standard simplex. I guess $n=2$ as an example will make things clearer.
n=2
You can embed a regular triangle in $\mathbb R^3$ with the unit vectors as corners: $P_1=(1,0,0),P_2=(0,1,0),P_3=(0,0,1)$. This triangle lives in the plane $x+y+z=1$. The center of the triangle is $C=\left(\frac13,\frac13,\frac13\right)$. So you get difference vectors like $C-P_1=\left(-\frac23,\frac13,\frac13\right)$ and $C-P_2=\left(\frac13,-\frac23,\frac13\right)$. But since you only care about the angle between them, you might as well use a multiple of these vectors to get rid of the fractions:
$$v_1=(-2,1,1)\qquad v_2=(1,-2,1)$$
Now you can use the dot product (which I write as $\left<*,*\right>$) to compute the cosine of the angle:
$$\cos\alpha=\frac{\left<v_1,v_2\right>}{\lVert v_1\rVert\cdot\lVert v_2\rVert}
= \frac{-2-2+1}{2^2+1+1}
= -\frac36 = -\frac12
$$
Higher dimensions
This generalizes to arbitrary dimensions, as follows:
$$
P_1 = (1,0,0,\dots,0) \qquad
C = \tfrac1{n+1}\cdot(1,1,1,\dots,1) \\
v_1 = (-n,1,1,\dots,1) \qquad
v_2 = (1,-n,1,\dots,1) \\
\left<v_1,v_2\right> = 2\cdot(-n)+(n-1)\cdot1 = -(n+1) \\
\lVert v_i\rVert^2 = 1\cdot n^2+n\cdot 1 = n(n+1) \\
\cos\alpha = \frac{-(n+1)}{n(n+1)} = -\frac1n
$$
as conjectured.
A: Yes, you can put $n$ vectors such as
$$ v_1 = ( n + \sqrt{1+n}, -1, -1, \ldots, -1  ),  $$
$$ v_2 = (-1, n + \sqrt{1+n}, -1,  \ldots, -1  ),  $$
and so on.
Then
$$ v_{n+1} = (-1 - \sqrt{1+n}, -1 - \sqrt{1+n}, -1 - \sqrt{1+n}, \ldots,-1 - \sqrt{1+n}, ).  $$
The center of the regular simplex is at the origin, the angles between pairs of vectors are $$ \arccos \, \frac{-1}{n}.  $$ 
Try it for $n=3$ and $n=8$ and $n=15,$ the square roots come out nice.
