Isomorphisms between the groups $U(10), U(5)$ and $\mathbb{Z}/4\mathbb{Z}$ I think its silly question but I have nt this in my mind at this time. 
Any one can help
How we can see $U(10) \overset{def}= \{1,3,7,9\}$, $\mathbb Z/4\mathbb Z \overset{def}= \{0,1,2,3\}$, $U(5) \overset{def}= \{1,2,3,4\}$ are isomorphic groups where $U(10)$ and $U(5)$ are groups under multiplication?
 A: General remark on group isomorphisms
Basically isomorphisms establish correspondance between the composition tables of two groups. If $(G,+)$ and $(H,+)$ are groups of the same size and we have an isomorphism
$$
\varphi:G\overset{\sim}{\longrightarrow}H
$$
then if we have elements $g_1,g_2\in G$ which composed gives $g_1+g_2=x\in G$ these are mapped via $\varphi$ to elements $h_1,h_2\in H$ that composed gives $h_1+h_2=y\in H$ in the way that
$$
g_1+g_2=x\overset{\varphi}{\longmapsto}y=h_1+h_2
$$
or using the standard language of homomorphisms
$$
\varphi(g_1+g_2)=\underbrace{\varphi(x)}_{y}=\underbrace{\varphi(g_1)}_{h_1}+\underbrace{\varphi(g_2)}_{h_2}
$$
Now, since $\varphi$ is injective the image of $G$ in $H$ is just as detailed as $G$ itself, since any distinct elements of $G$ maps to distinct elements of $H$, so the entire composition table of $G$ is found as a copy in $H$. Furthermore an isomorphism is surjective so that all of $H$ is described via the composition tables of $G$.
The order of an isomorphically mapped element
In particular the isomorphism $\varphi$ has to carry compositional properties from $G$ to $H$. This means in particular that the order of $g\in G$ corresponds to the order of $\varphi(g)\in H$.
If $g$ has order $k$ in $G$ this means that $g^k=\underbrace{g+g+...+g}_{k\mbox{ times}}=0$ and that this $k$ is minimal in this respect so that
$$
\{g,g^2,...,g^k\}
$$
is a subset of $k$ distinct elements of $G$. Now since
$$
\begin{align}
\varphi(g^m)&=\varphi(g+g+...+g)\\
&=\varphi(g)+\varphi(g)+...+\varphi(g)\\
&=\varphi(g)^m
\end{align}
$$
and since an injective map maps $k$ distinct elements to $k$ distinct elements this means that
$$
\{\varphi(g),\varphi(g)^2,...,\varphi(g)^k\}
$$
is a subset of $H$ containing $k$ distinct elements. And in particular
$$
g^k=0\overset{\varphi}{\longmapsto} 0=\varphi(g)^k
$$ showing that $\varphi(g)$ has order $k$ in $H$. This shows why you should map a generator to a generator...
The composition tables in your examples
$$
\newcommand{\red}{\color{red}}
\newcommand{\blue}{\color{blue}}
\newcommand{\Z}{\mathbb{Z}}
\begin{array}{c|c:c:c:c}
\Z_4&\ 0\ &\ \red1\ &\ \blue2\ &\ \red3\ \\
\hline
0&0&\red1&\blue2&\red3\\
\hdashline
\red1&\red1&\blue2&\red3&0\\
\hdashline
\blue2&\blue2&\red3&0&\red1\\
\hdashline
\red3&\red3&0&\red1&\blue2
\end{array}
\quad
\begin{array}{c|c:c:c:c}
U(10)&\ 1\ &\ \red3\ &\ \blue9\ &\ \red7\ \\
\hline
1&1&\red3&\blue9&\red7\\
\hdashline
\red3&\red3&\blue9&\red7&1\\
\hdashline
\blue9&\blue9&\red7&1&\red3\\
\hdashline
\red7&\red7&1&\red3&\blue9
\end{array}
\quad
\begin{array}{c|c:c:c:c}
U(5)&\ 1\ &\ \red2\ &\ \blue4\ &\ \red3\ \\
\hline
1&1&\red2&\blue4&\red3\\
\hdashline
\red2&\red2&\blue4&\red3&1\\
\hdashline
\blue4&\blue4&\red3&1&\red2\\
\hdashline
\red3&\red3&1&\red2&\blue4
\end{array}
$$
In these tables I used colours to distinguish the order of the elements.


*

*$\red{Red}$ denotes generators (order 4)

*$\blue{Blue}$ denotes the element of order 2 in each group

*$Black$ denotes the neutral element (order 1)

A: I think you can see $U(10) \cong U(5)$ by the Chinese remainder theorem.  We have
\begin{align*}
\left(\mathbb{Z}/10\mathbb{Z}\right)^\times &\stackrel{\sim}{\to} \left(\mathbb{Z}/2\mathbb{Z}\right)^\times \times \left(\mathbb{Z}/5\mathbb{Z}\right)^\times\\
1 &\mapsto (1,1)\\
3 &\mapsto (1,3)\\
7 &\mapsto (1,2)\\
9 &\mapsto (1,4)
\end{align*}
where the map is just reducing mod 2 and mod 5, respectively.  It's pretty clear that the trivial factor $\left(\mathbb{Z}/2\mathbb{Z}\right)^\times$ is unnecessary and this gives an isomorphism $U(10) \cong U(5)$.
Showing the isomorphism $U(5) \cong \mathbb{Z}/4\mathbb{Z}$ is trickier, but as others have said, you just need to send a generator to a generator.  In this case, you can check that $2$ is a generator for $U(5)$ since $2^2 = 4 \neq 1$.
