Suppose that $G$ is a simple $r$-regular graph with $n$ vertices. 
We say $H$ is a dominating set for $T$, if
for every vertex $v\in T$, we have $v\in H$ or there is a vertex $u\in H$ such that $vu\in E(G)$.


It is easy to see that if $T$ is an independent set for $G$, then there exists an independent
dominating set $H$ for $T$ such that $| T \cap H |=O(\log n)$.


Question: Suppose that $G$ is an $r$-regular graph with $r\neq 0$. If $T$ is an independent set for $G$, does there exist an independent
dominating set $H$ for $T$ such that $T \cap H =\emptyset$?


If the answer to Question is no, does there exist an independent
dominating set $H$ for $T$ such that $| T \cap H | =O(\log\log n)$?



A question about independent set in regular graphs
Clearly, if $G$ is $1$-regular or $2$-regular, and if $T$ is an independent set in $G$, then there is a maximal independent set $H$ in $G$ such that $T\cap H=\emptyset$.


Let $G$ be the $3$-regular graph of order $14$ with $V(G)=\{a,b,c,d,e,f,g,t,u,v,w,x,y,z\}$ 
and $E(G)=\{ab,bc,cd,de,ea,fb,fd,gc,ge,fg,tu,uv,vw,wx,xt,yu,yw,zx,zv,yz,at\}$.


Then $T=\{b,e,u,x\}$ is a maximal independent set in $G$, and if $H$ is any independent set in $G$ which is dominating for $T$, then $T\cap H\ne\emptyset$.


If $m\in\mathbb N$ then $mG$ is a $3$-regular graph of order $n=14m$ containing a maximal independent set $T'$ (with $|T'|=4m$) such that, if $H$ is any independent set in $mG$ which is dominating for $T'$, then $|T'\cap H|\ge m=\frac1{14}n$. This contradicts your "easy to see" claim about $|T\cap H|=O(\log n)$.


**P.S.** No doubt some assumption was omitted from the question. Connectedness is not enough, as the following construction shows.


**Theorem.** For each $m\in\mathbb N$ there is a connected cubic graph $G=(V,E)$ of order $|V|=n=50m$, and there is a maximal independent set $T\subseteq V$, with $|T|=14m$, such that, for any independent set $H\subseteq V$ which is dominating for $T$, we have $|T\cap H|\ge2m=\dfrac1{25}n$.


**Proof.** Take $6m$ vertices $a\_0,\ a\_1,\ a\_2,\dots,a\_{6m-1}$ and join them cyclically with edges $a\_0a\_1,\ a\_1a\_2,\dots,a\_{6m-1}a\_0$.


For each $j\in\{0,1,2,\dots,6m-1\}$, add vertices $x\_j,\ y\_j,\ z\_j,\ t\_j,\ u\_j,\ v\_j,\ w\_j$ and edges $x\_jy\_j,\ x\_jz\_j,\ y\_jt\_j,\ y\_ju\_j,\ z\_jv\_j,\ z\_jw\_j,\ t\_jv\_j,\ t\_jw\_j,\ u\_jv\_j,\ u\_jw\_j$.


For each $j\in\{0,1,2,\dots,6m-1\}\setminus\{1,4,7,\dots,6m-2\}$, add an edge $a\_jx\_j$.


For each $j\in\{1,4,7,\dots,6m-2\}$, add a vertex $b\_j$ and edges $a\_jb\_j,\ b\_jx\_j$.


Finally, join the vertices $b\_j$ in pairs with edges $b\_1b\_{3m+1},\ b\_4b\_{3m+4},\dots,b\_{3m-2}b\_{6m-2}$.


Now $T=\{a\_j:j=1,4,7,\dots,6m-2\}\cup\{y\_j:j=0,1,2,\dots,6m-1\}\cup\{z\_j:j=0,1,2,\dots,6m-1\}$ is a maximal independent set. If $H$ is any independent set which is dominating for $T$, then for each $j\in\{1,4,7,\dots,6m-2\},\ H$ must contain at least one of the vertices $a\_j,\ y\_j,\ z\_j,\ y\_{j-1},\ z\_{j-1},\ y\_{j+1},\ z\_{j+1}$, whence $|T\cap H|\ge2m$.


