Proving that $T$:$(x_1,...,x_n) \rightarrow (\frac {x_1+x_2}{2},\frac {x_2+x_3}{2},...,\frac {x_n+x_1}{2})$ leads to nonintegral components

The determinant of $T$, when nonzero, has to be an integer for $T$ to maintain integrality of vectors (proof: a simplex with integer vertices and minimal volume has to map to a finite union of such). This rules out odd $n$. Computation of the determinant can be accomplished using the formula for eigenvalues of circulant matrices, here $(1+\omega)/2$ with $\omega$ any $n$th root of unity.

For even $n$ the determinant of $T$ is $0$, but the map is not nilpotent, so there is potentially an integer sublattice to which the same argument applies (since there is only one $0$ eigenvalue). Indeed, $T$ preserves the $(n-1)$ dimensional subspace with $\sum (-1)^i x_i = 0$, and does not map any element of that subspace to $0$. This means that it is the invariant subspace of $T$ complementary to the $0$ eigenspace,and the determinant of $T$ restricted to that subspace is the product of the nonzero eigenvalues. The determinant again is not an integer.

First find the eigenvalues of $T$:

$Ty = \lambda y$ is equivalent to ${y_i + y_{i+1} \over 2} = \lambda y_i$ for all $i$, where $y_{n+1}$ is taken to mean $y_1$. This translates into $y_{i+1} = (2\lambda - 1)y_i$. Iterating $n$ times we have $(2\lambda - 1)^n y_i = y_i$. Since some $y_i$ is nonzero one has $(2\lambda - 1)^n = 1$ or equivalently $\lambda = {1 + \omega \over 2}$ where $\omega$ is any $n$th root of unity. Conversely, one can see that each such $\lambda$ is an eigenvector by using $y_1 = 1$ and $y_i = (2\lambda - 1)^{i-1} = w^{i-1}$ for each $i > 1$.

So the eigenvalues are of the form $\lambda_k = {\displaystyle{1 + e^{2\pi i (k-1) \over n} \over 2}}$ for $1 \leq k \leq n$ with some corresponding eigenvectors $v_k$. Any initial vector $x = (x_1,...,x_n)$ can be written in the form $c_1v_1 + .... + c_nv_n$ for some $c_i$. Since the $x_i$ are distinct and $v_1 = (1,...,1)$, we have that $c_i \neq 0$ for at least one $i > 0$. In the case where $n$ is even, $v_{{n \over 2} + 1} = (1,-1,1,-1,...,1,-1)$, so since the $x_i$ are distinct, if $n$ is even then $c_i \neq 0$ for some $i >1$, $i \neq {n \over 2} + 1$.

Next, note that $$T^k x = c_1 v_1 + \lambda_2^k c_2 v_2 + ... + \lambda_n^k c_n v_n$$ Since $|\lambda_i| < 1$ for all $i > 1$, as $k$ goes to infinity $T^k x$ converges to $c_1v_1$. By the last paragraph, there is always a $c_i \neq 0$ for which $i > 1$ and $\lambda_i^k \neq 0$, so since the $v_i$ are linearly independent for $i > 1$ it is not possible that for some $k$ one already has $T^k x = c_1 v_1$.

Next note that there is some $\epsilon > 0$ such that if $0 < ||w - c_1v_1|| < \epsilon$ then $w$ has noninteger components. Since $T^kx$ converges to $c_1v_1$ as $k$ goes to infinity and each $T^k x$ is not equal to $c_1v_1$, we have that for $k$ large enough $T^k x$ has noninteger components as needed.