Is the identity matrix the only matrix which is its own inverse?

$$\begin{pmatrix}\cos\theta & \sin\theta \\ \sin\theta &-\cos\theta\end{pmatrix}^{-1}=\begin{pmatrix}\cos\theta & \sin\theta \\ \sin\theta &-\cos\theta\end{pmatrix}$$ for any $\theta\in\mathbb{R}$.

You are looking for involutory matrices. To answer the question: no, there are other matrices that are their own inverses.

If $A^2=I$, then $(A-I)(A+I)=0$. Suppose that $J$ is the Jordan Canonical Form of $A$, then since $A=SJS^{-1}$, we have $$ \begin{align} 0 &=S^{-1}(A^2-I)S\\ &=S^{-1}AAS-S^{-1}IS\\ &=S^{-1}ASS^{-1}AS-I\\ &=J^2-I \end{align} $$ Thus, $J$ must be a diagonal matrix all of whose diagonal elements are $+1$ and $-1$.

Therefore, $A^2=I$ if and only if $A$ is diagonalizable and has eigenvalues $+1$ and $-1$.

Easy examples are diagonal matrices whose diagonal elements are $+1$ and $-1$, but any similar matrices will also work.