If $A$ is a non-square matrix with orthonormal columns, what is $A^+$? 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.
 A: 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.
A: The pseudoinverse $\rm A^+$ must satisfy $4$ conditions:


*

*$\rm A A^+ A = A$

*$\rm A^+ A A^+ = A^+$

*$\rm (A A^+)^* = A A^+$

*$\rm (A^+ A)^* = A^+ A$
Let us verify that $\color{blue}{\rm A^+ = A^\top}$. The first two conditions are satisfied because
$$\rm A A^+ A = A \underbrace{\left( A^\top A \right)}_{=I} = A$$
and
$$\rm A^+ A A^+ = \underbrace{\left( A^\top A \right)}_{=I} A^\top = A^\top = A^+$$
and the last two conditions are trivially satisfied if $\rm A$ is a real matrix.
A: 
If the real matrix $A$ has a left inverse, then $A^+=(A^TA)^{-1}A^T$.

Note that $A^TA$ being invertible is equivalent to $A$ having a left inverse.
If the columns of $A$ are orthonormal, then $A^TA=I$ and therefore $A^+=A^T$.
The initial statement is just a verification of the required properties for the pseudoinverse.

Suppose on the contrary that $A^+=A^T$. Consider $A=QR$, where $Q$ has orthonormal columns and $R$ has a right inverse (such a decomposition can be easily derived from the Gram-Schmidt algorithm). Then, by general theory,
$$
A^+=R^+Q^+=R^T(RR^T)^{-1}Q^T=A^T=R^TQ^T
$$
If we multiply by $Q$, we obtain $R^T(RR^T)^{-1}=R^T$ and therefore
$$
RR^T=I
$$
Thus $A$ is the product of a matrix with orthonormal columns and a matrix with orthonormal rows (the transpose of a matrix with orthonormal columns, if you prefer).
