Convex compact set in $\mathbb{R}^n$ where, given any point in it, the result of replacing two of its coordinates with their mean lies in the set. Let $X$ be a nonempty compact convex subset of $\mathbf{R}^n$.
Suppose this subset has the following property: for every $x = (x_1, \dots, x_n) \in X$, for every $1 \le i< j \le n$,
$$({x_1}, \ldots, {x_{i - 1}},
\frac{{x_i} + {x_j}}{2},
{x_{i + 1}},
\ldots,
{x_{j - 1}},
\frac{{x_i} + {x_j}}{2},
{x_{j + 1}},
\dots,
{x_n}
) \in X.$$
Is it true that there exists some $\lambda \in \mathbb{R}$ such that
$$(\lambda, \dots, \lambda) \in X?$$
One idea is that we can use the above property to get a sequence $(x, x', x'', \dots)$ where $x'$ is obtained by replacing two coordinates in $x$ with their average, and $x''$ in the same way, . . . Then we use sequential compactness to say that limit, call it $L$, also lies in $X$. Could we argue that every coordinate of $L$ is equal?
 A: Let $L:=\{\lambda(1,1,\cdots, 1):\lambda\in\mathbb R\}$ be the line we are interested. Since $X$ is compact, there is a point $x=(x_1, \cdots, x_n)$ such that $d(x, L)=\inf\{d(x,L):x\in X\}$.
Now the projection of $x$ to $L$ is $(x, \frac{(1, \cdots, 1)}{\sqrt n}) \frac{(1, \cdots, 1)}{\sqrt n}=\frac{\sum_{i=1}^n x_i}{n} (1, \cdots, 1)=\mathbb Ex (1, \cdots, 1)$ where $\mathbb Ex$ stands for the expectaton or averge of the coordinates of $x$, hence $d(x, L) = \|x-\frac{\sum_{i=1}^n x_i}{n} (1, \cdots, 1)\|=\sqrt{\sum_{i=1}^n (x_i-\mathbb Ex)^2}$.
For any $i, j$, after the operation, $\mathbb Ex$ is unchanged, and so is $(x_k-\mathbb Ex)^2$ for $k\not=i, j$. Meanwhile $(x_i-\mathbb Ex)^2+(x_j-\mathbb Ex)^2=x_i^2+x_j^2-2\mathbb Ex (x_i+x_j)+2(\mathbb Ex)^2$. None of the summands is changed except $x_i^2+x_j^2$ after averaging the two coordinates. And we have $2(\frac{x_i+x_j}{2})^2=\frac{(x_i+x_j)^2}{2}\le x_i^2+x_j^2$ where equality holds iff $x_i=x_j$. But $x$ minimizes the distance $d(x, L)$, hence $x_i=x_j$ must hold for all pairs.
In short, the continuous function $x\mapsto \sum_{i=1}^n(x_i-\mathbb Ex)^2$ must attain a minimum on $X$. If the minimum is not zero, it can be lowered by averaging two distinct coordinates.
Convexity is not used in the proof.
A: If $\ x\in X\ $ is your starting point, $\ \overline{x}=\frac{1}{n}\sum_\limits{i=1}^nx_i\ $,  and the procedure suggested in ronno 's comment is applied successively to obtain a sequence $\ x,x',\dots,x'^{\left.^{\large r}\right.},\dots\ $, then
$$
\big\|x'^{\left.^{\large r}\right.}-\overline{x}\mathbb{1}_{1\times n}\,\big\|^2\le\left(1-\frac{1}{2n}\right)^r\big\|x-\overline{x}\mathbb{1}_{1\times n}\,\big\|^2\ ,
$$
and so $\lim_\limits{r\rightarrow\infty}x'^{\left.^{\large r}\right.}=\overline{x}\mathbb{1}_{1\times n}\ $.
Let $\ x_a=\min_\limits{1\le i\le n}x_i\ $, $\ x_b=\max_\limits{1\le i\le n}x_i\ $, and $\ d=x_b-x_a\ $.  Then $\ \big|x_i-x_j\,\big|\le d\ $ for all $\ i,j\in\{1,2,\dots,n\}\ $, and
\begin{align}
\big\|x-\overline{x}\mathbb{1}_{1\times n}\,\big\|^2&=\frac{1}{n^2}\sum_{i=1}^n\left(nx_i-\sum_{j=1}^nx_j\right)^2\\
&=\frac{1}{n^2}\sum_{i=1}^n\left(\sum_{j=1}^n\big(x_i-x_j\big)\right)^2\\
&\le\frac{1}{n^2}\sum_{i=1}^n n^2d^2\\
&= nd^2
\end{align}
Let $\ T(x)\ $ be the vector obtained from $\ x\ $ by replacing $\ x_a\ $ and $\ x_b\ $ with $\ \frac{x_a+x_b}{2}\ $ (i.e. that obtained by adopting ronno's suggestion).   As pointed out in just a user's answer, $\ T(x)^\top\mathbb{1}_{1\times n}=x^\top\mathbb{1}_{1\times n}\ $, so
\begin{align}
\big\|x-\overline{x}\mathbb{1}_{1\times n}\,\big\|^2-\big\|T(x)-\overline{x}\mathbb{1}_{1\times n}\,\big\|^2&=\|x\|^2-\|T(x)\|^2\\
&=x_a^2+x_b^2-2\left(\frac{x_a+x_b}{2}\right)^2\\
&=\frac{\big(x_b-x_a\big)^2}{2}\\
&=\frac{d^2}{2}\ .
\end{align}
It follows from this, and the inequality $\ nd^2\ge\big\|x-\overline{x}\mathbb{1}_{1\times n}\,\big\|^2\ $ derived above, that
\begin{align}
\big\|T(x)-\overline{x}\mathbb{1}_{1\times n}\,\big\|^2&=\big\|x-\overline{x}\mathbb{1}_{1\times n}\,\big\|^2-\frac{d^2}{2}\\
&\le\left(1-\frac{1}{2n}\right)\big\|x-\overline{x}\mathbb{1}_{1\times n}\,\big\|^2\ .
\end{align}
And now, since $\ x'^{\left.^{\large r}\right.}=T^r(x)\ $, and $\ \overline{T(x)}=\overline{x}\ $, it follows by induction that
$$
\big\|x'^{\left.^{\large r}\right.}-\overline{x}\mathbb{1}_{1\times n}\,\big\|^2\le\left(1-\frac{1}{2n}\right)^r\big\|x-\overline{x}\mathbb{1}_{1\times n}\,\big\|^2\ ,
$$
as claimed above.
A: I am deeply confused. Do you mean subsets or subspaces?
For a standard nonempty compact convex subset just choose the rectangle
$$ 
X:=\{(x,y)\in R^2 | 2\leq x\leq3, 5 \leq 6 \leq 7\}
$$
The intersection with the line
$$
L:=\{\lambda (1,1), \lambda \ \in \mathbb{R}\} 
$$
is empty.
