If a matrix has orthonormal columns, they must be linearly independent, so $A^+ = (A^T A)^{−1} A^T$ . Also, the fact that its columns are orthonormal gives $A^T A = I$. Therefore,
$$A^+ = (A^T A)^{−1} A^T = (I)^{-1}A^T = A^T$$
Thus, $A^+ = A^T$. Am I correct? Thank you.
Best Answer
Problem statement
Start with a matrix $$A\in\mathbb{C}^{m\times n}$$ where $m>n$, and a valid statement for the pseudoinverse matrix $$ \mathbf{A}^{+} = \left( \mathbf{A}^{*} \mathbf{A} \right)^{-1} \mathbf{A}^{*} $$ We know (see links) that this matrix is a left inverse: $$ \mathbf{A}^{+} \mathbf{A} = \mathbf{I}_{n} \tag{1} $$
If we add the constrain that the column vectors of $\mathbf{A}$ are orthonormal, we also have $$ \mathbf{A}^{*} \mathbf{A} = \mathbf{I}_{n} \tag{2} $$
Conclusion
The results $(1)$ and $(2)$ suggest the identity $$ \mathbf{A}^{+} \mathbf{A} = \mathbf{I}_{n} = \mathbf{A}^{*} \mathbf{A} $$ from which we conclude that $$ \mathbf{A}^{+} = \mathbf{A}^{*} $$
Example
$$ \mathbf{A} = \frac{1}{\sqrt{2}} \left[ \begin{array}{cr} i & -1 \\ i & 1 \\ 0 & 0 \\ \end{array} \right], \qquad \mathbf{A}^{*} = \frac{1}{\sqrt{2}} \left[ \begin{array}{rrc} -i & -i & 0 \\ -1 & 1 & 0 \\ \end{array} \right] $$
$$ \mathbf{A}^{+} \mathbf{A} = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right] = \mathbf{I}_{2}, \qquad \mathbf{A} \mathbf{A}^{+} = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{array} \right] \ne \mathbf{I}_{3} $$
$$ \mathbf{A}^{*} \mathbf{A} = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right] = \mathbf{I}_{2}, \qquad \mathbf{A} \mathbf{A}^{*} = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{array} \right] \ne \mathbf{I}_{3} $$
Background reading
Categorize the pseudoinverse matrix in terms of left and right inverses: generalized inverse of a matrix and convergence for singular matrix, What forms does the Moore-Penrose inverse take under systems with full rank, full column rank, and full row rank?
General properties of the pseudoinverse matrix: Moore–Penrose pseudo-inverse Reference.