Points fixed by the isotropy group of the restricted Poincaré group in Minkowski space What points does the isotropy group $SO(3)$ fix in the Minkowskian case? For instance, in oriented Euclidean 3-space the isotropy group $SO(3)$ of the Euclidean group $E(3)$ fixes the origin. What is the case in Minkowski 4-space with the isotropy group $SO(3)$ for timelike directions?
On request: Consider the action of the Poincaré group on the space of future directed timelike curves. This action is transitive. That means we are considering a single orbit in Minkowski space, not particular orbits with fixed origin of the Lorentz subgroup of the Poincaré group. We are picking as orbit the space of forward cone hyperboloids and the stabilizer that corresponds to each of these hyperboloids in Minkowski space is $SO(3)$(or $SU(2)$ if we consider the doble cover of the Poincare group as usual in QFT).
So it would seem that the whole set of points of the manifold is fixed by the isotropy group in the transitive action of the Poincaré group on Minkowski space, not just the origin of the rotation, which would be the case just for the action of the Lorentz group acting on a point in Minkowski space.
In fact no element of the Poincaré group can act on Minkowski space fixing just a point.
 A: Let $(L, \langle \cdot , \cdot \rangle)$ be a 4-dimensional Lorentzian space (of signature (3,1)); let $P_+$ denote the orientation-preserving subgroup in the Poincare group $P$ of $(L, \langle \cdot , \cdot \rangle)$:
$$
P_+\cong {\mathbb R}^4\rtimes SO(3,1).
$$
Consider a Lie subgroup $H< P_+$ isomorphic to $SO(3)$. I will use the following neat argument due to Cartan:
Lemma. Let $K$ be a compact group of affine isometries of a Hilbert space ${\mathcal H}$, preserving a nonempty closed convex subset $C\subset {\mathcal H}$. Then $K$ fixes a point in $C$. 
Proof. Consider an orbit $Kx\subset C$. Since $K$ is compact, so is the orbit $Kx$. Let $b\in {\mathcal H}$ denote the center  of mass of the orbit $Kx$. Since $C$ is convex and $Kx\subset C$, it follows that $b\in C$. Since the center of mass is unique, it is fixed by $K$. qed.  
This lemma has several corollaries. For instance, if $K$ is a compact group of affine transformations of a $n$-dimensional vector space $V$ (or even an infinite dimensional topological vector space as long as it admits 
an inner product), then $K$ acts by isometries of some Euclidean metric on 
$V$. To prove this, consider the action of $K$ on the space of all positive-definite bilinear forms on $V$ (the action is via the pull-back). This space can be seen as a convex subspace in the space of all $n\times n$-symmetric matrices; the latter has an invariant inner product given by $tr(A^TB)$. Hence, by the above lemma, $K$ has a fixed point in the space of positive-definite bilinear products on $V$, hence, preserves some Euclidean metric on $V$. Applying the lemma one more time, to now the action of $K$ on $V$ itself, we conclude that $K$ has a fixed point in $V$. 
Corollary. Let $K$ be a compact group of affine transvormations of a Hilbert space ${\mathcal H}$, preserving a nonempty closed convex subset $C\subset {\mathcal H}$. Then $K$ fixes a point in $C$. 
I now go back to your question. Since $H\cong SO(3)$ is compact, the group $H$ fixes a point in $L$. I will declare this point to be the origin $o$ in $L$, i.e. $H$ is contained in the stabilizer $P_o$ of  the origin in $P_+$, i.e. in the Lorentz group
$$
P_o\cong SO(3,1). 
$$
Now, $H$ acts on $L$ via linear transformations. The next thing to observe is that $H$ fixes a time-like vector $v\in L$, i.e. a vector such that $\langle v, v \rangle <0$. Indeed, $P_o$ preserves the set
$$
\{x\in L: \langle x, x\rangle\le -1\}.  
$$
This set is not convex but it is the disjoint union of two convex connected components $C_+, C_-$ (future and past directions). Since $SO(3)$ is connected, it preserves both $C_+$ and $C_-$. Since $H$ preserves $C_+$, by applying the Corollary, we conclude that $H$ fixes a vector $v\in C_+$. 
Hence, we obtain:
Lemma. 1. Every subgroup $H\cong SO(3)$ in $P_o$ fixes pointwise a time-like line $\ell\subset L$. (This is the line spanned by the vector $v$.)


*Every subgroup $H\cong SO(3)$ in $P_+$ fixes pointwise an affine time-like line $\ell\subset L$. 


As far as I can tell, you are asking "what is the fixed-point set of $H$ in $L$ provided that $H< P_+$ is isomorphic to $SO(3)$?" 
[Recall that the fixed-point set of a group $H$ acting on a set $S$ is the subset of $S$ consisting of all $x\in S$ such that $h(x)=x$ for all $h\in H$.]  
This question has the answer: The affine line $\ell$ defined above. 
Let's prove this claim. First of all, as I explained above, we can choose the origin in the affine space $L$ to be a point $o\in \ell$ (fixed by $H$); then $H$ acts linearly on $L$ with respect to this choice of the origin. (I think, this change of the origin is the source of confusion in your comments.) 
It is clear from the definition of $H$ that it fixes all points of $\ell$. It also preserves the Lorentzian orthogonal complement $\ell^\perp$ of the line $\ell$. The bilinear form $B=\langle \cdot , \cdot \rangle$ restricts to a positive definite bilinear form on $\ell^\perp$ and $H$ acts on $\ell^\perp$  as the full group of orientation-preserving orthogonal transformations of this form. Hence, $H$ has no nonzero fixed vectors in $\ell^\perp$. Since the fixed point set of $H$ is a linear subspace of $L$, which intersects $\ell^\perp$ only at the origin, it follows that this fixed point set is 1-dimensional, hence, equals to the line $\ell$. qed 
