Involution centralizer of perfect group with quaternion Sylow 2-subgroups

Solution 1:

Let $P$ be generated by $x$ of order $2^n > 4$ and $y$ of order $4$, and let $z = x^{2^{n-2}}$ be a power of $x$ of order 4. Then the two classes of $Q_8$ in $P$ are represented by $\langle z,y \rangle$ and $\langle z, xy \rangle$.

From what you say, you know that both of these subgroups are normalized in $X$ by subgroups isomorphic to ${\rm SL}(2,3)$. That means that $z$ is conjugate in $X$ to both $y$ and to $xy$. I don't think that's possible in a solvable group $X$.

Assume that $X$ is solvable, so it has a chief series with elementary abelian factors. So $z \in M \setminus N$ for one of these factors $M/N$. If $z$ is conjugate to both $y$ and $xy$, then we have $y, xy \in M \setminus N$. But then $x \in M$ and so $x^2 \in N$ and hence $z \in N$, contradiction.

ADDED: For your smaller question if, for conjugates $u$ and $v$ of $t$, $x=uv$ is nontrivial and centralizes $t$, then $x$ has odd order and both $t$ and $u$ lie in the normalizer $N$ of $\langle x \rangle$ and hence are conjugate in $N$, but that is impossible because $t$ centralizes and $u$ inverts $x$.