Let $G \subset \mathrm{GL}_n(k)$ be a connected affine algebraic group over a field $k$ with the following property: for any two distinct elements $g,h \in G$ there exists a vector $x \in k^n, x\neq 0$ such that $g\cdot x = h \cdot x$.
Is it true then that there exist vectors $x_0,y_0 \in k^n, x_0,y_0 \neq 0$ such that for all $g \in G$, $g\cdot x_0 = y_0$?
update: as is rightfully remarked in the comments, if one considers commuting elements (e.g. powers $g, g^2, g^3, \ldots$) then one notices that $y_0=x_0$.
another observation: an equivalent condition would be that all $g \in G$ fix a vector (indeed, $(gh^{-1})x=x$ implies $g$ and $h$ agree on $x$). Then the question can be rephrased as follows: assume all $g \in G$ fix a vector. Is it true then that they have a common fixed vector?
I will post a counterexample to the following statement, equivalent to the inital one: if $G \subset \mathrm{GL}_n$ is an affine connected group, and every element $g \in G$ fixes a vector, then all $g \in G$ fix one common vector. Thanks to Kostya Tolmachev for pointing this counterexample out.
Take the representation of $SL_2$ on degree 2 homogeneous vectors in variables x,y: this is a vector space with basis $x^2, xy, y^2$ and $\left( \begin{array}{cc} a & b \\ c & d\\ \end{array} \right)$ acts by $x \mapsto ax + by, y \mapsto cx + dy$.
This representation of $SL_2$ is known to be irreducible, hence no common fixed vectors. Suffices to check that any element of a Borel subgroup has eigenvalue 1 in the representation, so pick $g \in SL_2$ upper-triangular matrix in the standard representation, and write down what it looks like in the representation described above: $$ \left( \begin{array}{ccc} a^2 & ac & c^2 \\ 0 & 1 & 2\frac{c}{a} \\ 0 & 0 & \frac{1}{a^2} \end{array} \right) $$ Then it is clear that it has an eigenvalue 1.