Let $X$ be a smooth projective variety with an action of $\mathbb{C}^{*}$. Let us suppose that the set $X^{\mathbb{C}^{*}}$ is finite. For $x \in X^{\mathbb{C}^{*}}$, let $A_{x}$ denote the attractor (under our $\mathbb{C}^{*}$-action) to $x$ and let $R_{x}$ denote the repellent. Take $x_{1}, x_{2} \in X^{\mathbb{C}^{*}}$. Suppose that $x_{2} \in \bar{A}_{x_{1}}$, where $\bar{A}_{x_{1}}$ is the closure of $A_{x_{1}}$.
Is that true that $x_{1} \in \bar{R}_{x_{2}}$?
Counterexample: Take the action of $\mathbb C^*$ on $\mathbf P^2(\mathbb C)$ given by $$ t\cdot[a_0:a_1:a_2]=[a_0:ta_1:t^2a_2]. $$ There are three fixed points $p_0$, $p_1$, $p_2$ where $p_0$ is a sink and $p_2$ is a source. There are two orbits having $p_1$ in its closure, namely the lines $L_0$, $L_2$ through $p_0,p_1$ and $p_1,p_2$, respectively.
Now let $X$ be the blow up of $\mathbf P^2(\mathbb C)$ in $p_1$ with exceptional divisor $E$. Then $X$ has four fixed points $p_0,p_1',p_1'',p_2$ where $L_0\cap E=\{p_1'\}$ and $L_2\cap E=\{p_1''\}$. Moreover, $L_0,E,L_2$ are the only orbit closures not containing both $p_0$ and $p_2$. They are organized in a chain.
The fixed point $x_1:=p_0$ is still a sink. So, the closure of its attracting set is all of $X$. In particular, it contains $x_2:=p_1''$. On the other side, the closure of the repelling set of $p_1''$ is $E$ which does not contain $x_1$.
