Is it necessary $A^TA =I$ if $AA^T=I$ where A is a square matrix? Suppose if $AA^T=I$ where $A$ is a square matrix, is it necessary that $A^TA=I$?

I tried to google it but didn't find anything confirmatory.

In the same way if $AA^{-1}=I$, is it necessary that $A^{-1}A=I$?

My attempt it as follows:-

$$A=\begin{bmatrix}
a&amp;b&amp;c\\
d&amp;e&amp;f\\
g&amp;h&amp;i
\end{bmatrix}$$

$$AA^T=\begin{bmatrix}
a&amp;b&amp;c\\
d&amp;e&amp;f\\
g&amp;h&amp;i
\end{bmatrix}\begin{bmatrix}
a&amp;d&amp;g\\
b&amp;e&amp;h\\
c&amp;f&amp;i
\end{bmatrix}$$

$$AA^T=\begin{bmatrix}
a^2+b^2+c^2&amp;ad+be+cf&amp;ag+bh+ci\\
da+eb+fc&amp;d^2+e^2+f^2&amp;dg+eh+fi\\
ga+hb+ic&amp;gd+he+if&amp;g^2+h^2+i^2
\end{bmatrix}$$

$$a^2+b^2+c^2=1$$
$$d^2+e^2+f^2=1$$
$$g^2+h^2+i^2=1$$
$$ad+be+cf=0$$
$$ag+bh+ci=0$$
$$da+eb+fc=0$$
$$dg+eh+fi=0$$
$$ga+hb+ic=0$$
$$gd+he+if=0$$

Now if we calculate $A^TA$, it would be 

$$A^TA=\begin{bmatrix}
a^2+d^2+g^2&amp;ab+de+gh&amp;agc+df+gi\\
ba+ed+hg&amp;b^2+e^2+h^2&amp;bc+ef+hi\\
ca+fd+ig&amp;cb+ef+ih&amp;c^2+f^2+i^2
\end{bmatrix}$$

Now can one say $a^2+d^2+g^2=1$ from the previous equations? I think not necessary. So by this exercise I doubt $A^TA=I$

I am not getting any idea for $AA^{-1}=I$
Yes, because inverses are unique.
In addition to all this $I=I^T=(AA^T)^T =A^T(A^T)^T=A^TA $.