EDIT: I was wrong, and the actual answer is that there IS such a mapping. We need to specify it on the boundary, and check that the winding (i.e. image of the boundary in $H^n (D_n \ e)$ is 0). The image of the identity map has index 1.

Now, consider the simplex of dimension 6. Note that it has 6 five-dimensional facets, 6*5/2 = 15 four-dimensional facets, 6*5*4/6= 20 three-dimensional facets.

Note that gcd(6, 15, 20) = 6+15-20 = 1. It is possible to construct such a map that it is identity away from the small neighborhoods of the centers of 5, 4, 3 - dimensional facets. Near the centers of the 5 and 4-dimensional facets we define the map as follows:

Denote the center of the facets as $c$. Away from the circle of radius $R<<1$ it is identity. Inside it it is identity plus vector $(e-c) f(r)$, where $r$ is a distance to $c$, and $f(r) = (1+\varepsilon)(1-\frac{r}{R})$. $\varepsilon << 1$.

Each such facet will decrease winding number by $1$.

Near the 3-dimensional facets, the following reverse construction works:

Fix some $R_2 < R$. Out of the circle of radius $R$ the map F is identity). In the annulus $R_2 < r < R$ it is $F(x) = c + (x-c)\frac{r - R_2}{R-R_2}$. In the inner circle the map is $c - (x-c)\sin(\pi \frac{R_2-r}{R_2}) + (e-c)f(r)$.

Here, $f(r) = (1+\varepsilon)(1-\frac{r}{R_2})$

Each such facet will lower the winding by $1$.

Now, the resulting map will have winding $0$, hence can be prolonged to the interior of the simplex.

I think that it exists if and only if gcd($C_n^k$) over all $1<k<n$ equals 1.

Previous answer (WRONG):

I think you can prove this fact in a following way (I'd like to denote by $D_n$ the simplex of dimension $n$, so $D(n+1)$ in your notation) consider the image of a boundary. It is some homological cycle $s \in H^{n-1} (D_n \ e) \simeq \mathbb{Z}$.

It is clear that any two symmetric maps from $D_n$ to itself can be transformed to each other by a homotopy in a class of symmetric maps. One can also arrange this homotopy in such a way that only the images of the $n-1$-dimensional facets ever pass through $e$.

Then, it is clear that this homology element changes by some multiple of $n$ on each such occasion (because facets pass through $e$ simultaneously due to symmetry condition), and by exactly $n$ if the passing is transverse.

For the identity mapping, $s = 1$, so by the argument above for any mapping $s = 1+kn \neq 0$. Hence, the image of $D_n$ should intersect $e$ (because if it won't intersect $e$ the element $s$ would be homologous to $0$.~~