I don't know where you found your claim, but it seems that the distance induced by the Frobenius norm between any two orthogonal matrices can be any real number.
Because:
$$
\begin{align}
\|A - B\|^2 &= \mathrm{tr} \left( (A-B)^t (A-B) \right) \\
&= \mathrm{tr} \left( (A^t -B^t) (A-B) \right) \\
&= \mathrm{tr} (A^tA -A^tB - B^tA + B^tB) \\
&= \mathrm{tr} (2I) - \mathrm{tr}(2A^tB) \\
&= 2n - 2\mathrm{tr}(A^tB)
\end{align}
$$
The last but one equality is due to the fact that $A^tA = B^tB = I$ and $(B^tA)^t = A^tB$ and a matrix and its transpose have the same trace.
Now, take $A = I$ and we've got
$$
\| I - B\|^2 = 2n - 2\mathrm{tr}(B)
$$
for any orthogonal matrix $B$. So, if you take as $B$ the family of orthogonal matrices
$$
\begin{pmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{pmatrix} \ ,
$$
$\theta \in [0, 2\pi]$, their traces $\mathrm{tr} (B) = 2\cos\theta$ can be any real number between $-2$ and $2$. So, their Frobenius distances to the unit matrix $I$ can be any real number from $0$ to $\sqrt{8}$.
Let's consider what happens when we multiply $Q^TQ$.
The $ij$ entry with $i\neq j$ is the dot product of row $i$ of $Q^T$ (i.e., column $i$ of $Q$) with column $j$ of $Q$. Since the columns of $Q$ are orthogonal, this is $0$.
The $ii$ entry is the dot product of column $i$ of $Q$ with itself, which is always $1$ because the columns of $Q$ are normal.
So $Q^TQ$ has $1$s down the diagonal and $0$s elsewhere; i.e., it is the identity matrix.
Thus $Q^T$ is the inverse of $Q$.
Best Answer
If $A^t = A^{-1}$, then taking inverses of both sides, we have $(A^{t})^{-1} = A = (A^t)^t$.