Let $0 \neq S \in \mathfrak{so}(3)$ be a non-zero skew-symmetric matrix, and let $R \in SO(3)$. The claim is that, indeed, $SR$ is skew-symmetric if and only if $R = I_3$ or $R$ is the rotation by $\pi$ about the axis $\ker(S)$; since $SI = S$ is necessarily skew-symmetric, we only need to consider the case where $R \neq I_3$. Observe that
$$
(SR)^T+SR = R^TS^T+SR = -R^TS+SR =-R^T(S-RSR),
$$
so that $SR$ is skew-symmetric if and only if $S=RSR$.
Before continuing, let me check that $\ker(S)$ is indeed $1$-dimensional. Observe that, since $S$ is skew-symmetric, that $\ker(S)$ and $\ker(S)^\perp$ are both $S$-invariant. Now, since $S \neq 0$, fix a unit vector $0 \neq x \in \ker(S)^\perp$. Then, since $S$ is skew-symmetric, $x$ is orthogonal to $0 \neq Sx \in \ker(S)^\perp$, so that $\ker(S)^\perp$ is at least two-dimensional and hence $\ker(S)$ is at most $1$-dimensional. However, $S$ cannot be invertible, for $\operatorname{set}(S) = \operatorname{det}(S^T) = \operatorname{det}(-S) = -\operatorname{det}(S)$, so that, indeed, $\operatorname{det}(S) = 0$. Thus, $\ker(S)$ is, in fact, $1$-dimensional.
For future convenience, observe that if $0 \neq x \in \ker(S)^\perp$ is a unit vector, then for $y := \tfrac{1}{\|Sx\|}Sx$ and $0 \neq z \in \ker(S)$ any unit vector, $\{x,y\}$ is an orthonormal basis for $\ker(S)^\perp$, $\ker(S) = \mathbb{R}z$, and $\beta = \{x,y,z\}$ is an orthonormal basis for $\mathbb{R}^3$. Indeed, let’s see how $S$ looks in terms of this basis. On the one hand, we already have that $Sz = 0$ and that $Sx = \alpha y$ for $\alpha = \|Sx\|$. On the other hand, by the same argument as above, $y$ is orthogonal to $0 \neq Sy \in \ker(S)^\perp$, so that
$$
Sy = \langle x, Sy \rangle = -\langle Sx, y \rangle = -\langle \alpha y, y \rangle = -\alpha.
$$
Thus,
$$
Sx = \alpha y, \quad Sy = -\alpha x, \quad Sz = 0,
$$
for $\alpha := \|Sx\| \neq 0$.
Suppose, on the one hand, that $SR$ is skew-symmetric, and hence that $S=RSR$. We first need to show that $\ker(S) = \ker(R-I_3)$, so assume otherwise. Let $z \in \ker(S)$ be a unit vector. Fix a unit vector $x \in \ker(R-I_3)$, so that $\ker(R-I_3) = \mathbb{R}x$. Then,
$$
z = z_0 + cx
$$
for $z_0 \in \ker(R-I_3)^\perp$ and $c = \langle x,z\rangle$; by assumption, $z_0 \neq 0$. Then
$$
0 = Sz = RSRz = RSR(z_0 + cx) = RS(Rz_0 + cx),
$$
so that $Rz_0 + cx \in \ker(S)$, i.e., $Rz_0 + cx = az$ for some $a \in \mathbb{R}$. But then,
$$
Rz_0 + cx = az = az_0 + acx,
$$
so that $Rz_0 = az_0$ and $cx = acx$. Since $z_0$ is non-zero, it is therefore a real eigenvector for $R$, forcing $a = -1$ and hence $c = 0$. Thus, $R$ is the rotation by $\pi$ about the axis $\ker(R-I_3) = \mathbb{R}x$, and $\ker(S) \subset \ker(R-I_3)^\perp = \ker(R+I_3)$.
But now, starting with the unit vectors $x \in \ker(R-I_3) \subset \ker(S)^\perp$ and $z \in \ker(S)$, we can construct, as above, an orthonormal basis $\{x,y,z\}$ of $\mathbb{R}^3$, where $y = \tfrac{1}{\|Sx\|}Sx$, such that
$$
Sx = \alpha y, \quad Sy = -\alpha x, \quad Sz = 0
$$
for $\alpha = \|Sx\| \neq 0$. But then, in particular, since $y \in \ker(R-I_3)^\perp = \ker(R+I_3)$,
$$
-\alpha x = Sy = RSRy = -RSy = \alpha R x = \alpha x,
$$
contradicting $\alpha \neq 0$.
So, as we have seen, $\ker(S) = \ker(R-I_3)$, so again, choose a unit vector $x \in \ker(S)^\perp = \ker(R-I_3)^\perp$ and construct an orthonormal basis $\{x,y,z\}$ of $\mathbb{R}^3$ such that
$$
Sx = \alpha y, \quad Sy = -\alpha x, \quad Sz = 0,
$$
for $\alpha = \|Sx\| \neq 0$. Since $S$ is skew-symmetric, $\ker(S)^\perp = \ker(R-I_3)^\perp$ is $R$-invariant, so that, in particular, $Rx \in \ker(S)^\perp$. But then, since $SR$ is skew-symmetric,
$$
0 = \langle x, SR x \rangle = -\langle Sx, Rx \rangle = -\alpha \langle y, Rx \rangle,
$$
so that $Rx \in \ker(S)^\perp$ is orthogonal to $y$, and hence $Rx = cx$ for some $c \in \mathbb{R}$, i.e., $x \in \ker(S)^\perp = \ker(R-I_3)^\perp$ is a non-zero real eigenvector for $R$. Thus, necessarily, $c = -1$, so that, indeed, $R$ is the rotation by $\pi$ about the axis $\ker(R-I_3) = \ker(S)$.
Suppose, on the other hand, that $R$ is the rotation by $\pi$ about the axis $\ker(S) = \ker(R-I_3)$. Let $z \in \mathbb{R}^3$, so that $z = z_0 + z_1$ for $z_0 \in \ker(S) = \ker(R-I_3)$ and $z_1 \in \ker(S)^\perp = \ker(R+I_3)$. Then, since $S$ is skew-symmetric, $\ker(S)^\perp = \ker(R+I_3)$ is $S$-invariant, and hence
$$
RSRv = RSR(v_0 + v_1) = RS(v_0-v_1) = -RSv_1 = Sv_1 = Sv_0 + Sv_1 = Sv,
$$
as required.
"how to obtain skew-symmetrical part of rotation matrix skew(R), knowing its symmetrical part sym(R)?"
I will assume you are talking about rotations in $\mathbb{R}^3$. First note that a rotation about an axis of angle $\theta$ and a rotation about the same axis of angle $-\theta$ have the same symmetric parts. But apart from this ambiguity, it is possible to reconstruct the skew-symmetric part of a rotation matrix (up to a sign) by only knowing the symmetric part. Thus one can reconstruct the whole rotation matrix (apart from this ambiguity above) from the symmetric part.
Proof:
Case 1: $\operatorname{sym}(R)-I \neq 0$.
By Rodrigues formula, $\operatorname{sym}(R)-I = (1-\operatorname{cos}(\theta))S^2(v)$. For a square matrix $A$, you can define its norm by $||A||^2 = \frac{1}{2} Tr(A^TA)$. Then by normalizing $\operatorname{sym}(R)-I$, one gets rid of $1-\operatorname{cos}(\theta)$, and gets $S^2(v)$.
The kernel of $S^2(v)$ is the axis of rotation. It remains to recover $\operatorname{sin}(\theta)$, up to a sign. Well, in $\operatorname{sym}(R)-I$, the factor multiplying $S^2(v)$ is $1-\operatorname{cos}(\theta)$, so $\operatorname{cos}(\theta)$ is known, and from it we can get $\operatorname{sin}(\theta)$ up to a sign.
Case 2: $\operatorname{sym}(R)-I=0$
By Rodrigues formula, this corresponds to $\operatorname{cos}(\theta) = 1$, i.e. to $R = I$. This finishes the proof.
Best Answer
Rodrigues formula is:
$v′=(\cos \theta)v + (\sin \theta)b \times v + (1− \cos \theta) b (b \cdot v)$
Using the fact that:
$b \times (b \times v)=b(b \cdot v)−v(b \cdot b)$
Using $\|b\| = 1$:
$b \times (b \times v)=b(b \cdot v)−v$
$(1− \cos \theta) b \times (b \times v) = (1− \cos \theta)b (b \cdot v) − v + (\cos \theta) v$
Now we need to accomodate those terms in Rodrigues formula:
$v′= v + (\sin \theta)b \times v + (1− \cos \theta ) b (b \cdot v) − v + (\cos \theta) v $
Replacing:
$v′= v + (\sin \theta)b \times v + (1− \cos \theta) b \times (b \times v)$
Now you can apply the matrix identities:
$K v = b \times v$
$K^2 v = b \times (b \times v)$
To get:
$v′= v + (\sin \theta)K v + (1− \cos \theta) K^2 v$
Finally, you can factor out $v$ to get the matrix $R$:
$v′= R v$
$R = I + (\sin \theta)K + (1− \cos \theta) K^2$