The correct answer is $\boxed{\left[ {\begin{array}{*{20}{c}} {\frac{1}{4}}&0&0&0 \ 0&{\frac{1}{4}}&0&0 \ 0&0&{\frac{1}{2}}&0 \ 0&0&0&{\frac{1}{2}} \end{array}} \right]}$.
An orthogonal matrix is a square matrix whose columns and rows are orthonormal vectors. This means that the dot product of any two columns or rows is equal to 1, and the dot product of any column with itself is equal to 1.
The inverse of an orthogonal matrix is also an orthogonal matrix. This means that if $A$ is an orthogonal matrix, then $A^{-1}$ is also an orthogonal matrix.
The inverse of an orthogonal matrix can be found using the formula $A^{-1} = \frac{1}{|A|}A^T$, where $|A|$ is the determinant of $A$.
In this case, $A$ is given by the following matrix:
$$A = \left[ {\begin{array}{*{20}{c}} 1&1&1&1 \ 1&1&{ – 1}&{ – 1} \ 1&{ – 1}&0&0 \ 0&0&1&{ – 1} \end{array}} \right]$$
The determinant of $A$ can be found using the formula $|A| = \det \left[ {\begin{array}{*{20}{c}} 1&1&1&1 \ 1&1&{ – 1}&{ – 1} \ 1&{ – 1}&0&0 \ 0&0&1&{ – 1} \end{array}} \right] = -4$.
Therefore, the inverse of $A$ is given by the following matrix:
$$A^{-1} = \frac{1}{|A|}A^T = \frac{1}{-4}\left[ {\begin{array}{*{20}{c}} {\frac{1}{4}}&0&0&0 \ 0&{\frac{1}{4}}&0&0 \ 0&0&{\frac{1}{2}}&0 \ 0&0&0&{\frac{1}{2}} \end{array}} \right]$$