One can easily check that for a given vector $\gamma$, $U\gamma$ for all the $U$'s which are rotation matrices defines a sphere with the Euclidean metric.
For a given matrix $A\in M_{m\times n}(\mathbb{R})$, with Frobenius norm and based on singular valued decomposition one can easily see that $UA$ generates the set $S=\{UA|U\in O(m)\}$ where we have $$\forall U\in O(m), \|UA\|_F=\|A\|_F.$$
It appears to me this set should be a sphere and its dimension should somehow be related to the rank of $A$ but I am unable to draw anything. Is there any known result in this direction or can somebody give me some hints?
Thank you for this interesting question. I'd like to answer it as follows. I'm going to use some facts about topology, groups, operations, orbits, etc. Hopefully, you're familiar with that.
First, let us review your initial observation, namely that for $v \in \mathbb R^m \setminus \{0\}$, the set $O(m)v$ is $S_{|v|}^{m - 1}$, the sphere of radius $|v|$ in $\mathbb R^m$ centered at the origin. Of course, this follows from the construction of $O(m)$. Now, consider the map $$\Phi: O(m) \rightarrow S_{|v|}^{m -1} \subseteq \mathbb R^m,A \mapsto Av.$$ In other words, $\Psi$ maps $O(m)$ to the orbit of $v$ under $O(m)$ in $\mathbb R^m$. This map is continuous and surjective. Let $$F_v := \left\{ A \in O(m) \mid| Av = v \right\}$$ be the stabilizer of $v$ in $O(m)$. Then, by using some standard facts about groups, operations, and orbits, $\Phi$ induces a bijections $$\overline\Phi : O(m) / F_v \rightarrow S_{|v|}^{m -1}.$$ By using some facts about topological groups, $\overline\Phi$ is continuous, and, by using the compactness of $O(m)$ (and some more topological technicalities), even a homeomorphism. In words, we have shown that the Orbit $O(m)v$ is homeomorphic to the group $O(m)$ modulo the stabilizer $F_v$.
Now, if we put $$v = e_1 := (1, 0, \ldots, 0)^T$$ and write $I_k$ for the k by k identity matrix, then we find $$F_{e_1} = \left\{ I_1 \right\} \times O(m-1) \subseteq O(m).$$ In other words, $F_{e_1}$ is "$O(m-1)$ in the lower right hand corner of $O(m)$". In particular $F_{e_1} \simeq O(m-1)$, here $\simeq$ denotes isomorphism.
Note that, for general $v$, $F_v$ and $F_{e_1}$ are conjugate in $O(m)$. More specifically, there is a $B \in O(m)$ such that $Bv = |v|e_1$ (check that). With that $B$, we have $$F_v = B^{-1}F_{e_1}B.$$ That's a general fact about stabilizers. We have also used that $F_{e_1} = F_{|v|e_1}$ (check that). If we denote right conjugation by $B$ on $O(m)$ by $\Gamma_B$, i.e. $$\Gamma_B:O(m) \rightarrow O(m), A \mapsto B^{-1}AB,$$ then the map $\Gamma_B$ is extremely well behaved - it's a bijection, a homeomorphism, a diffeomorphism, an automorphism. From all these nice properties, together with the fact $$\Gamma_B(F_{e_1}) = F_v$$ (see above), we get that $\Gamma_B$ induces a homeomorphism $$O(m)/F_{e_1} \simeq O(m)/F_v.$$ This shows that for any $v$ we get the same quotient up to homeomorphism. So, using the particular form of $F_{e_1}$ from above, we can introduce the seductive notation $$O(M)/O(m-1) := O(m)/\left( \left\{ I_1 \right\} \times O(m-1)\right) \simeq O(m)/F_v$$ for general $v$. Putting together all the maps and identifications we have constructed, we find $$O(m)/O(m-1) \simeq S^{m-1},$$ where $S^{m-1}$ is the unit sphere in $\mathbb R^m$.
Of course, all of this is just a more detailed description of your observation $O(m)v = S_{|v|}^{m-1}$. Now we come to the interesting part where we apply the above considerations to your actual question.
So we fix an $n \in \mathbb N$ and consider the operation of $O(m)$ on matrix space $\mathbb R^{m\times n}$ via left multiplication. We pick a matrix $$W = \left(v^{(1)}, \ldots, v^{(n)}\right) \in \mathbb R^{m\times n}$$ where we denote the columns of $W$ by $v^{(j)}$. Let us also assume that $W$ has no zero columns, because zero columns don't add anything interesting to the problem. Now, the orbit of $W$ under $O(m)$ is of course $O(m)W$, and we want to describe its topological structure. To this effect, we want to proceed as before and use the fact that "the orbit is homeomorphic to the group modulo the stabilizer" (see above). So, we have to determine the stabilizer $$F_W := \left\{ A \in O(M) \mid| AW = W \right\}.$$ To this end, we put $$\mathcal U := \left< v^{(1)}, \ldots, v^{(n)} \right >_{\mathbb R} \subseteq \mathbb R^m,$$ the linear span of the $v^{(j)}$ in $\mathbb R^m$. If we put $$r := rank(W),$$ then we have $$\dim \mathcal U = r.$$ We have the orthogonal decomposition $$\mathbb R^m = \mathcal U \oplus \mathcal U^\perp$$ with $$ \dim \mathcal U^\perp = m - r.$$ With all this, it's "very easy to see" that for $A \in O(m)$ we have $\left( AW = W \right)$ iff $\left( \forall j: Av^{(j)} = v^{(j)} \right)$ iff ($A$ fixes $\mathcal U$ pointwise) iff $\left( A \in \left\{ id_{\mathcal U} \right\} \times O(\mathcal U^\perp) \subseteq O(m) \right)$, where $id_{\mathcal U}$ is the identity mapping on $\mathcal U$ and $O(\mathcal U^\perp)$ denotes the orthogonal group on the vector space $\mathcal U^\perp$. Again, it's "easy to see" that $$ \left\{ id_{\mathcal U} \right\} \times O(\mathcal U^\perp) \simeq \left\{ I_r \right\} \times O(m - r)$$ (notation see above) where the isomorphism is given by conjugation in $O(m)$. To summarize, we have just shown that $$F_W \simeq \left\{ I_r \right\} \times O(m-r)$$ by conjugation in $O(m)$. From all of this, we get in exactly the same way as above the following homeomorphisms. $$O(m) / O(m - r) := O(M)/\left(\left\{ I_r \right\} \times O(m-r)\right) \simeq O(m) / F_W \simeq O(m)W$$
So far so good - now we know that $O(m)W$ is $O(m) / O(m - r)$. But now we're at our wits' end because we have no idea how $O(m) / O(m - r)$ looks like. But thankfully, we have Google and Wikipedia, and after some trial and error, we hit upon
http://en.wikipedia.org/wiki/Stiefel_manifold
and find that $O(m) / O(m - r)$ is in fact a Stiefel manifold! From what I gather from that page, Stiefel manifolds are quite intricate and intriguing geometric objects. So it's no wonder you have difficulties drawing them.