There are infinitely many figures like that.
Let $p(\theta)$ be any smooth periodic function of period $2\pi$. It is known that
the necessary and sufficient condition for such a function $p(\theta)$ to be the
support function of a convex set $K$ is
$$p(\theta) + p''(\theta) > 0 \quad\text{ for all }\;\theta$$
The boundary of the convex set $K$ is given by the parametrization:
$$\begin{cases}
x &= p(\theta)\cos\theta - p'(\theta)\sin\theta,\\
y &= p(\theta)\sin\theta + p'(\theta) \cos\theta
\end{cases}
$$
If you pick a function $p(\theta)$ which satisfies an additional constraint
$$p(\theta)^2 + p\left(\theta + \frac{\pi}{2}\right)^2 = 1$$
then for every $\theta$, the pair of tangent lines at $\theta$ and $\theta + \frac{\pi}{2}$ intersect at some point on the unit circle. Since they are perpendicular to each other, the angle subtended by $K$ at that intersection will be a right angle.
As an example, pick any $\alpha \in [0,1)$ and consider the function
$\displaystyle\;p(\theta) = \sqrt{\frac{1+\alpha \cos(2\theta)}{2}}$.
It is easy to check
$\displaystyle\;p(\theta) + p''(\theta) = \frac{1-\alpha^2}{\sqrt{2}(1 + \alpha\cos(2\theta))^{3/2}} > 0$ for all $\theta$.
This give us a convex set bounded by the curve
$$\begin{cases}
x(t) &= \frac{1+\alpha}{\sqrt{2}}\frac{\cos(t)}{\sqrt{1 + \alpha\cos(2\theta)}}\\
y(t) &= \frac{1-\alpha}{\sqrt{2}}\frac{\sin(t)}{\sqrt{1 + \alpha\cos(2\theta)}}
\end{cases}$$
With help of an CAS, one can simplify this to the ellipse
$\displaystyle\;\frac{x^2}{1+\alpha} + \frac{y^2}{1-\alpha} = \frac12$.
This is equivalent to the director circle mentioned by others in comment.
For a more complicated example, one can try something like
$\displaystyle p(\theta) = \sqrt{\frac{1+\beta\cos(6\theta)}{2}}$
where $\beta \in [0,\frac{1}{17})$ instead. This will give you a figure
that looks like a hexagon with the edge curved outward.