Rank $L$ spherical harmonic tensor as a $2L+1$ dimensional Cartesian vector? Rank two Cartesian tensors can be decomposed into $L=0,1,2$ spin like things  3x3=1+3+5.
But the second equation below does not transform like a "tensor", it looks more like a vector transform in five dimensional Cartesian coordinate.
$$\mathfrak{D}^{+}(R)V_{i}\mathfrak{D}(R) = \sum_{j=1,2,3}R_{ij}V_{j}\tag{1} $$
(three dimensional vector.)
$$\mathfrak{D}(R)T_{i}^{(2)}\mathfrak{D}^{+}(R) = \sum_{j=-2,-1,0,1,2}\mathfrak{D}_{ij}^{(2)}T_{j}^{(2)}\tag{2}$$
(rank two spherical harmonic tensor.)
$$ \mathfrak{B}^{+}(F)V_{i}\mathfrak{B}(F) = \sum_{j=1,2,3,4,5}F_{ij}V_{j}\tag{3}$$ (a vector transform in 5 dimensional Cartesian space.) 
$(2)$ and $(3)$ above looks the same.
My question is can 3d spherical tensors be regarded as high dimensional vectors? Are there some examples?
 A: Vectors in $\mathbb R^5$ transform under the action of $SO(5)$, instead all the representations you are considering (for every value of $l=0,1,2,…$) are irreducible representations of $SO(3)$, so what you suppose is not possible. 
However, barring the standard spherical harmonics intepretation,  the $l=2$ representation has a precise geometric meaning as the action of $SO(3)$ on an irreducible space of 2nd order real tensors in $\mathbb R^3 \otimes \mathbb R^3$ with dimension $5$. I go to illustrate it.  
Since you are considering integer values of the spin, the arising
$SU(2)$ representation are also representations of $SO(3)$. More
precisely the matrices in the right-hand side of (1) and (2) corresponds
to unitary irreducible representations of $SO(3)$:
$$U(R) T^{(l)}_m U(R)^\dagger = \sum_{|m'| \leq l} D^{(l)}_{mm'}(R)T_{m'}$$
These matrices work in invariant and irreducible subspace of complex tensor
representations of $SO(3)$. The maps $SO(3) \ni R \mapsto D^{(l)}(R)$,
varying $l=0,1,2,\ldots$ label all complex unitary irreducible
representations of $SO(3)$ (as a byproduct one sees that these
representations are necessarily finite dimensional  all that is a
consequence of Peter-Weyl theorem).
Actually one can say more: all complex irreducible representations
of $SO(3)$ are obtained form these unitary irreducible
representations via a  transformation of the form: $$D^{(l)}(R) \to V
D^{(l)}(R) V^{-1}\qquad (1)$$ for suitable non singular $(2l+1) \times
(2l+1)$ matrices $V$ independent form $R$.
Let us consider the action of $SO(3)$ on $\mathbb R^3\otimes \mathbb
R^3$. If $SO(3) \ni R \to R : \mathbb R \to \mathbb R$ is the
fundamental representation, the tensor representation on $\mathbb
R^3\otimes \mathbb R^3$ is $SO(3) \ni R \mapsto R \otimes R$. However
$\mathbb R^3\otimes \mathbb R^3$ is not irreducible, but it consists of
a direct sum of $3$ real irreducible subspaces:
$$\mathbb R^3\otimes \mathbb R^3 = \mathbb S_1 \oplus \mathbb S_2 \oplus
\mathbb S_3$$
$\mathbb S_1$ is the subspace of scalars: the $3\times 3$ real matrices
of the form $sI$, $\mathbb  S_2$ is the space of antisymmetric  $3\times
3$ real matrices, $\mathbb S_3$ is the space of traceless symmetric
$3\times 3$ real matrices. If $M \in \mathbb R^3\otimes \mathbb R^3$,
its decomposition in the three spaces above is:
$$M = \frac{1}{3}(tr M) I + \frac{M-M^t}{2} + \left(\frac{M+M^t}{2}-
\frac{1}{3}(tr M)I\right)$$
These three spaces are separately irreducible and invariant under
$SO(3)$ and transform with corresponding real irreducible
representations of $SO(3)$ that I indicate with $SO(3) \ni R \mapsto
L^{(i)}(R)$.
NOTE: If $M_{ij}= -M_{ji}$, $t_k :=
\sum_{i,j=1}^3\epsilon_{kij}M_{ij}$ transform as a real $3$-vector under
$SO(3)$ (the situation would be different if $SO(3)$ were replaced by
$O(3)$). So actually $\mathbb S_2$ is a vector representation of
$SO(3)$. In components of elements of $\mathbb S_2$, the representation
$L^{(2)}$
 is nothing but the fundamental representation $SO(3) \ni R \mapsto R$.
If we complexificate these spaces, $\mathbb S_i +i \mathbb S_i$, every
$SO(3) \ni R \mapsto L^{(i)}(R)$  becomes a complex irreducible
representation of $SO(3)$. So:
up to a transformation like in (1) for some fixed matrix $V^{(i)}$,
each representation  $L^{(i)}$ must coincide with one of the $D^{(l)}$
ones.
The first irreducible space $\mathbb S_1$ has real dimension $1$ so that
$\mathbb S_1 + i \mathbb S_1$ has complex dimension $1$. It means that
$L^{(1)}= D^{(0)}$. It is evident that $V^{(0)}=1$.
The second irreducible space $\mathbb S_2$ has real dimension $3$ so
that $\mathbb S_2 + i \mathbb S_2$ has complex dimension $3$. It means
that $L^{(2)}$ coincides to $D^{(1)}$ up to a similitude transformation
as in (1). The form of $V^{(1)}$ can be found in the literature, is the
matrix used to pass from spherical to Cartesian components.
The third irreducible space $\mathbb S_3$ has real dimension $5$ so that
$\mathbb S_3 + i \mathbb S_3$ has complex dimension $5$. It means that
$L^{(3)}$ coincides to $D^{(2)}$ up to a similitude transformation as in
(1). I do not know the explicit expression of  $V^{(2)}$.
