The strategy is: (1) show $A$ has eigenvalue $1$, (2) extend the corresponding eigenvector to an orthonormal basis, (3) show the matrix in this basis is a $2 \times 2$ rotation block plus a fixed axis.

**Step 1: $A$ has eigenvalue $1$.**

[claim:Eigenvalue One Exists] If $A \in \mathrm{SO}_3(\mathbb{R})$, then $\det(A - I) = 0$. [/claim]

[proof] Since $A^T A = I$ and $\det A = 1$: \begin{align*} \det(A - I) &= \det(A - AA^T) = \det A \det(I - A^T) = \det(I - A^T) \\ &= \det(I - A^T)^T = \det(I - A) = (-1)^3 \det(A - I) = -\det(A - I). \end{align*} So $\det(A - I) = -\det(A - I)$, giving $\det(A - I) = 0$. Hence $1$ is an eigenvalue. [/proof]

**Step 2: Construct an orthonormal basis.**

Let $\mathbf{v}$ be a unit eigenvector with $A\mathbf{v} = \mathbf{v}$. Extend $\mathbf{v}$ to an orthonormal basis $\{\mathbf{a}, \mathbf{b}, \mathbf{v}\}$ of $\mathbb{R}^3$.

**Step 3: Determine the matrix in the new basis.**

Let $P$ be the change-of-basis matrix with columns $\mathbf{a}, \mathbf{b}, \mathbf{v}$. Then $P \in O_3(\mathbb{R})$ and: \begin{align*}

A' = P^T A P = \begin{pmatrix} a & b & 0 \\ c & d & 0 \\ 0 & 0 & 1 \end{pmatrix}, \end{align*} since the third column of $A'$ is the image of $\mathbf{v}$, which is $\mathbf{v}$ itself (coordinates $(0, 0, 1)^T$ in this basis). Since $A' \in \mathrm{SO}_3(\mathbb{R})$ (conjugation preserves orthogonality and determinant), the upper-left $2 \times 2$ block $Q = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ satisfies $Q^T Q = I_2$ and $\det Q = 1$. This forces $Q \in \mathrm{SO}_2(\mathbb{R})$, so:

\begin{align*}

Q = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \end{align*} for some $\theta \in [0, 2\pi)$. Therefore $A$ is conjugate to a rotation by angle $\theta$ about the axis $\mathbf{v}$.