Let $\|\cdot\|\_F$ denote the Fröbenius norm on the set of $d\times d$ matrices. By restriction this induces a metric on $SO(n)$. 


Let's make an observation.
Since $X\in SO(n)$ is a rotation matrix then it is an isometry hence if $\lambda$ is an eigenvalue of $A$ with corresponding eigenevector $x$ we have that
$$
\|x\|=\|Ax\|=\|\lambda x\|= |\lambda| \|x\| 
\,\Rightarrow\,
|\lambda|=1.
$$
Therefore, we get the crude bound
$$
\begin{aligned}
\sup\_{X, Y \in SO(n)} \|X-Y\|\_F 
\leq &
\sup\_{X,Y \in SO(n)} \|X\|\_F + \|Y\|\_F
\\= &
\sup\_{X, Y \in SO(n)} \sqrt{
\sum\_{i=1}^n \lambda\_i(X)
}
+
\sqrt{
\sum\_{i=1}^n \lambda\_i(Y)
} 
\\= & 2\sqrt{n}
,
\end{aligned}
$$
where I use $\lambda\_i(X)$ to emphasize the $i^{th}$ eigenvalue of $X$. 


However, here are my two issues with this bound:


1. It is not specific to $SO(n)$ and applies to any set of linear isometries of $\mathbb{R}^n$,
2. It is clearly crude since it entirely disregards the distance between $X$ and $Y$ and only looks at their "norm" individually..


Is a sharp(er?) estimate for
$$
\sup\_{X,Y \in SO(n)} \|X-Y\|\_F,
$$
known? Specifically, can we bound this quantity by $1$?



Bound between distance between Rotation Matrices
**First Point: The bound isn't sharp**


Consider the case where $n=2$. The every matrix in $SO(n)$ is of the form
$$
A\_{\theta} \triangleq \begin{pmatrix}
cos(\theta) & -sin(\theta)\\
sin(\theta) & cos(\theta),
\end{pmatrix}
$$
for some $\theta \in [0,2\pi]$ (note: fun easy proof of compactness of $SO(2)$). In particular,
$$
\|A\_0 - A\_{\frac{\pi}{2}}\|\_F =
\left\|\begin{pmatrix}
1 & -1\\
1 & 1,
\end{pmatrix}\right\|\_F= \sqrt{4} = 2.
$$
So $2\sqrt{2}$ is not sharpe.


**Second point: $1$ cannot be achieved**
This also shows that $1$ cannot be achieved if $SO(n)$ is metrized by the Fröbenius norm, since we just got $2$...


**Third point/question: $1$ can *maybe* be Achieved if we instead consider the Spectral Norm**
Recall that the spectral (or Operator norm) of a $n\times n$ matrix is given by
$$
\|X\|\_{\infty} = \max\_{i=1,\dots,n} |\sigma\_i(A)|.
$$
Therefore, by your remark on the eigenvalues of any $A \in SO(n)$ we have that
$$
\sup\_{X,Y \in So(n)}\, \|X-Y\|\_{\infty} \leq 2.
$$
However, a quick computation shows that
$$
\|A\_0 - A\_{\frac{\pi}{2}}\|\_{\infty} = \sqrt{2}>1.
$$
So $1$ cannot be achied..


**Suggestion:** If you're willing to take any metric induced by a norm on the set of $n\times n$ matrices then I would just use
$$
\|X-Y\|\_n' := \frac1{2\sqrt{n}} \|X-Y\|\_F.
$$
Note that it generates the same topology on $SO(n)$ since all norms are equivalent on finite-dimensional normed spaces...
So, if you can use this, then your bound will give you a metric induced by a norm which is uniformly bounded by $1$ on $SO(n)$! 


*Hopefully this works for you.*


