= Solution
\b[True.] Use the <odd-dimensional orthogonal determinant splitting>:
$$
\boxed{\Phi:O(3)\to SO(3),\qquad \Phi(A)=(\det A)A.}
$$
Scalar matrices commute with every matrix, so $\Phi(AB)=\Phi(A)\Phi(B)$. Its output is orthogonal and
$$
\det\Phi(A)=(\det A)^3\det A=(\det A)^4=1.
$$
Every $B\in SO(3)$ satisfies $\Phi(B)=B$, proving surjectivity. The equation $\Phi(A)=I$ forces $A=(\det A)I$, whence
$$
\boxed{\ker\Phi=\{I,-I\},\qquad |\ker\Phi|=2.}
$$
In odd dimension $-I$ has <determinant> $-1$, which is the sign needed in this construction.
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