Strict convexity of the Frobenius ball but not of the operator ball

Idea. A ball is strictly convex if the midpoint of two distinct points of the sphere lies strictly inside the ball. Euclidean balls satisfy this; the operator-norm ball looks like a "cube" in the diagonal directions and has flat faces.

Step 1: the operator norm. Let $D=\operatorname{diag}(1,0,\dots,0)$, so $D\ne I$ as $n\ge2$. For $t\in[0,1]$,
$$tI+(1-t)D=\operatorname{diag}(1,t,\dots,t),$$
whose operator norm is $\max(1,t)=1$ (norm of a diagonal matrix = largest absolute value of an entry). So the whole segment $[D,I]$ lies on the sphere $\{\|A\|=1\}$ and has positive length $\|I-D\|=1$.

Step 2: the Frobenius norm. Let $X\ne Y$ with $\|X\|_F=\|Y\|_F=1$. By the parallelogram law, which holds because $\|\cdot\|_F$ comes from an inner product,
$$\Big\|\frac{X+Y}2\Big\|_F^2=\frac{\|X\|_F^2+\|Y\|_F^2}2-\Big\|\frac{X-Y}2\Big\|_F^2=1-\frac{\|X-Y\|_F^2}4<1.$$
So the midpoint lies strictly inside the ball; no segment of positive length lies in the sphere (its midpoint would be on it).

Remark. For $n=1$ both norms coincide with $|a|$ and the unit ball is $[-1,1]$, whose sphere $\{\pm1\}$ contains no segment, which is why $n\ge2$ is needed. The lack of strict convexity of the operator ball is linked to its extreme points being just $O(n)$ (see "Extreme points"): the flat faces consist of non-extreme points.