First of all in algebra people tend not to call them "creators", we tend to say generators if I am correct. Secondly, to check a group is not cyclic you would have to verify that none of its members generates the entire group. In your case it can be done extremely easy by inspection: Note that $$\langle 3\rangle =\langle 3,3^2=1\rangle$$ and the same hold for $5$ and $7$.
To create an isomorphism just map the generator of one of your groups to the generator of the other. That is, $\varphi:U_5\rightarrow U_{10}$ will be defined as $\varphi(2)=3$, everything else will be determined from here:$$\varphi(4)=\varphi(2\cdot 2)=\varphi(2)\varphi(2)=3\cdot 3=9$$$$\varphi(3)=\varphi(2\cdot 4)=\varphi(2)\varphi(4)=3\cdot 9=2$$$$\varphi(1)=\varphi(4\cdot 4)=\varphi(4)\varphi(4)=9\cdot 9=1$$
Note the last line was not even necessary since in a homomorphism identities are mapped to identities.
Hope this helps.
The quaternion group is indeed the minimal counter-example. Clearly, any group of orders $1,2,3,5$ or $7$ is cyclic, and the two (non-isomorphic) groups of order $4$ are:
$\Bbb Z_4$ and $\Bbb Z_2 \times \Bbb Z_2$.
There are likewise just two non-isomorphic groups of order $6$:
$\Bbb Z_6$ and $S_3 \cong \Bbb Z_3 \rtimes \Bbb Z_2$ (this is the only possible non-trivial semi-direct product of these two groups, since:
$0 \mapsto 1_{\Bbb Z_3}\\1 \mapsto (x \mapsto -x)$
is the sole non-trivial homomorphism $\Bbb Z_2 \to \text{Aut}(\Bbb Z_3)$).
It is convenient to use this formulation of a(n internal) semi-direct product:
$G = NH$, where $N,H$ are subgroups of $G$, and $N \lhd G$.
$N \cap H = \{e_G\}$.
The problem with obtaining $Q_8$ as a semi-direct product of two proper subgroups, is that we must have either $|H|$ or $|N|$ equal to $2$. But the only subgroup of order $2$ in $Q_8$ is $\{1,-1\}$, which is a subgroup of every subgroup of $Q_8$ of order $4$:
$\langle i\rangle = \{1,-1,i,-i\}\\\langle j\rangle = \{1,-1,j,-j\}\\\langle k\rangle = \{1,-1,k,-k\}$
Note that all $6$ elements of order $4$ lie in one of these $3$ subgroups.
Thus the condition $N \cap H = \{e_G\}$ (which is $=\{1\}$ in this case) is impossible to satisfy.
Best Answer
There are only two groups of order four: (1) the cyclic group and (2) the Klein group.
As all elements of $U(8)$ are of order $2$, $U(8)$ is indeed isomorphic as a group to the Klein group.
The key argument is that there is no other groups of order four than the two mentioned above.