On a projective plane $\pi_+:=\pi\cup l_\infty$ in which $\pi$ is the ordinary plane and $l_\infty$ is the infinity line, a projective transformation $f$ maps a circle $C\subset\pi$ to itself, then does $f$ necessarily map $C$'s centre $O$ to itself?
If $f(C)$ is only an ellipse, then I believe $f(O)$ isn't necessarily the centre of $f(C)$. However I'm unsure about the circle case.
On
Let $\alpha$ be a projective plane with line at infinity $w$, $\Gamma,\Psi$ be center conics (possible $\Gamma=\Psi$) in $\alpha$ and let $\Pi$ be a plane homography (projectivity) of the plane $\alpha$ sending $\Gamma$ into $\Psi$. Then
$\Pi$ sends the center of $\Gamma$ in the center of $\Psi$ if and only if $\Pi$ is an affinity.
Recall that a plane homography is an affinity if and only it fixes the line at infinity, that's $\Pi(w)=w$.
If we identify $\Gamma$ and $\Psi$ with their associated polarity of the plane $\alpha$, to say that $\Pi$ sends $\Gamma$ in $\Psi$ is equivalent to require $\Pi\circ\Gamma=\Psi\circ\Pi$. Since the line at infinity $w$ is the polar of the center of a center conic, $W=\Gamma(w)$ is the center of $\Gamma$ and $W'=\Psi(w)$ is the center of $\Psi$. Since \begin{align} \Pi(W) &=\Pi\circ\Gamma(w)\\ &=\Psi\circ\Pi(w) \end{align} we have \begin{align} \Pi(W)=W' &\iff \Psi\circ\Pi(w)=W'\\ &\iff \Psi\circ\Pi(w)=\Psi(w)\\ &\iff\Pi(w)=w \end{align} hence $\Pi(W)=W'$ if and only if $\Pi$ is an affinity.
In the complex plane $$ z\mapsto \frac{1+(1+i)z}{(1-i)+z} $$ is a map bringing the set $\|z\|=1$ into itself and $0$ into $-\frac{1+i}{2}$. It can be regarded as a projective transformation on $\mathbb{P}^2(\mathbb{R})$, showing that circle centers are not preserved. See also Cayley transform and circle inversion. For instance, if $\Gamma_1,\Gamma_2$ are two orthogonal circles in the plane, the circle inversion with respect to $\Gamma_2$ brings $\Gamma_1$ into itself, but the center of $\Gamma_1$ is not mapped into itself: