How can find the center of SmallGroup($64$,$41$)=$(\mathbb Z_{16}\rtimes\mathbb Z_2)\rtimes \mathbb Z_2$ with the following presentation without using GAP?
$$\langle F_1,F_2\mid F_2^2, F_1^4, F_1F_2F_1^2F_2F_1^{-1}F_2F_1^2F_2, (F_1F_2)^2F_1^{-1}(F_1^{-1}F_2)^2F_1, (F_2F_1F_2F_1^{-1})^2(F_2F_1^{-1})^2(F_2F_1)^2\rangle$$
Added by Derek Holt: Here is a presentation of the group that might be more amenable to hand calculation.
$$\langle x,y,z \mid x^{16}=y^2=z^2=1, x^y=x^9, x^z=x^{-1}y, y^z=x^8y \rangle.$$
With Holt's presentation: (we write $h^a=a^{-1}ha$, hence $h^{ab}=(h^a)^b$).
Centralizer of $x$ contains $\langle x\rangle$, but it does not contain $y$ and $z$. Can it contain $yz$?
$$x^{yz}=(x^y)^z=(x^9)^z=(x^z)^9=(x^{-1}y)^9\neq x$$ since there will be a factor of $y$ in the expansion [this can be seen by explicit calculation: $$(x^{-1}y).(x^{-1}y)=x^{-1}.(y^{-1}x^{-1}y)=x^{-1}.x^{-9}=x^{-10}=x^{6}.$$ Hence $(x^{-1}y)^2=x^6$ and one can find $$(x^{-1}y)^9=[(x^{-1}y)^2]^4.(x^{-1}y)=x^{6.4}x^{-1}y=x^iy\neq x$$ Thus, centralizer of $x$ is $\langle x\rangle$, and hence $Z(G)$ is a subgroup of this group of order $16$; it can not be $\langle x\rangle$.
$z^{-1}x^2z=(x^{-1}y)^2=x^6\neq x^2.$ Hence $Z(G)\neq \langle x^2\rangle$.
Also, $z^{-1}x^4z=(x^{-1}y)^4=x^6.x^6=x^{12}\neq x^4$.
You can check that $x^8$ is in the center and it generates center: $Z(G)=\langle x^8\rangle\cong C_2$.