The following claim is a well-known consequence of the Whitney-Graustein theorem:
Claim. It does not exist $H\colon\mathbb{S}^1\times[0,1]\overset{C^1}{\rightarrow}\mathbb{R}^2$ such that for all $t\in [0,1]$, $H(\cdot,t)\colon\mathbb{S}^1\rightarrow\mathbb{R}^2$ is an immersion, $H(\cdot,0)=(\cos(2\pi\cdot),\sin(2\pi\cdot))$ and $H(\cdot,1)=(\cos(2\pi \cdot),-\sin(2\pi\cdot))$.
In other words, it is impossible to perform a circle eversion in the plane, namely it is impossible to continuously and regularly change the orientation of the circle while sticking to the plane.
However, I want to illustrate that it is possible to realize the circle eversion in the $3$-dimensional space.
The idea is to thicken the circle into a cylinder, perform a $\pi$-twist on the cylinder in order to put its inside out and finally to retract the everted cylinder onto its equatorial circle.
My main concern is to graphically represent the above process using a mathematical software, e.g. SageMath. I tried in vain to write down explicit formulas for it and here I am stuck. Please note that the following homotopy did not seem to do any good:
$$\forall x\in\mathbb{S}^1\times [-1,1],\forall t\in [0,1],H(x,t)=\frac{x}{\|x\|^{2t}}.$$
Any enlightenment will be greatly appreciated!
As pointed out by Mike Miller in the comments, the process I described is unnecessarily complex. In order to perform the circle eversion in the $3$-dimensional space, it suffices to assume that the circle lies in the $(xOy)$ plane and then gradually operate a rotation of angle $\pi$ and axe $x$ on it. This is achieved by the following : $$H\colon\left\{\begin{array}{ccc} \mathbb{S}^1\times[0,1]& \rightarrow & \mathbb{R}^3\\ (x,t) & \mapsto & (\cos(2\pi x),\cos(\pi t)\sin(2\pi x),\sin(\pi t)\sin(2\pi x)) \end{array}\right..$$
As a bonus, here are some pictures on how to unravel a figure-eight knot:
And just for fun, here is the unraveling in real life: