Continuous map on the simplex (applicable to fair division) Let $g$ be a continuous function from the unit simplex $D(n)$ (with $n$ vertices) into itself, that leaves invariant its vertices, and such that $g$ is not onto: to fix ideas say that $g(D(n))$ does not contain $e=((1/n),(1/n),...,(1/n))$.
I believe that g cannot satisfy the following symmetry property: 
Let $F_{i}$ be the face of $D(n)$ where the $i$-th coordinate is zero; and $s^{ij}$ be the permutation of $[n]$ exchanging $i$ and $j$. Then $g$ commutes with $s^{ij}$ on $F_{i}\cup F_{j}: g(s^{ij}(x))=s^{ij}(g(x))$.
This is trivial for $n=2$ and easy for $n=3$. By the symmetry assumption, the winding number in $D(3)$ of the image by $g$ of $F_{i}$ (viewed as a path between two vertices, taken clock-wise) with respect to $e$ does not depend on $i$, and is equal to $1/3+k$ (even though not all three symmetries are rotations). Hence, the image by g of $\Delta (3)$, the frontier of $D(3)$ viewed as a circular path, has a non zero winding number. This is impossible because $\Delta (3)$ is contractible in $D(3)$ whereas its image is not, as it avoids $e$. 
    For higher dimensions, the notion of a winding number is more complicated and may or may not help; perhaps an argument can be made without this notion.
 A: 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 $0$.~~
A: To Taras Banakh and IJL
Apologies to both of you for expressing the symmetry property in this cryptic way. Here is a clearer way, I hope:
Let $F_{i}$ be the face of $D(n)$ where the $i$-th coordinate is zero; and $s^{ij}$ be the permutation of $[n]$ exchanging $i$ and $j$. Then $g$ commutes with $s^{ij}$ on $F_{i}∪F_{j}$: $g(s^{ij}(x))=s^{ij}(g(x))$.
In particular if $f(0,1,0)$ is the vertex of the second coordinate, then $f(1,0,0)$ is the vertex of the first.
A: The question is nice! provided that in the hypotheses you actually assume that the barycenter is not in the image.
Your other question, where you only assume that g is not onto, admits easy counterexamples:
it is easy (eg by means of a partition of the unity) to make on your (n-1)-dimensional simplex Dn a smooth vector field V such that
- V vanishes at each point of each face of dimension <=n-3;
- V is transverse to each hyperface F (face of dimension n-2) at every point x interior to F, and enters Dn at this point.
Then, you consider the action of the symmetric group S_n on the simplex, and you change V to the sum of its images through all the symmetric trannsformations. This way, you have moreover that V is S_n-invariant.
Then, the time 1 of the flow of V is a counterexample to your question.
