I know that the answer is yes. Are the reciprocal of the eigenvalues of A the eigenvalues for $A^{-1}$? If the eigenvalues for A are $3$ and $2$, would the eigenvalues for $A^{-1}$ be $1/3$ and $1/2$?
What does it mean when an orthonormal eigenbasis of A is also an orthonormal eigenbasis of $A^{-1}$?
Let the orthogonal decomposition of $A$ be denoted as $$A = Q^TDQ$$ where $Q$ is orthogonal and $D$ is diagonal. Since A is invertible, $$A^{-1} = {\left(Q^TDQ\right)}^{-1} = Q^{-1}D^{-1}{(Q^T)}^{-1} = Q^TD^{-1}Q$$ as the transpose of an orthogonal matrix is is its inverse. Now recall that the inverse of a diagonal matrix is simply a diagonal matrix with the reciprocals of each diagonal element (assuming no zeros). For instance $$\begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix} ^{-1} = \begin{pmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{3} \end{pmatrix}$$ Since $A$ was invertible there will be no zeros on the diagonal of $D,$ so $A^{-1} $ is orthogonally diagonalizable with the same orthogonal matrix, $Q$ as $A$.
Note that in a orthogonal diagonalization, the diagonal elements in $D$ are you eigenvalues and $Q$ is an orthogonal eigenbasis. For more information, read up on similarity transformations.