Let $m$ be coprime to $n$. Then $A:=m U(n)$ is a group under multiplication modulo $mn$, and $A\approx U(n)$.
Indeed, $A$ is closed under multiplication (if $m\mid x$ and $m\mid y$, then $m\mid xy$ and $m\mid xy\bmod{mn}$).
Consider he map $\phi\colon A\to U(n)$, $x\mapsto x\bmod n$ (note that this can be defined because if $x=my\in m U(n)$, then $\gcd(x,n)=\gcd(m,n)\gcd(y,n)=1$).
Then clearly $\phi(xy)=\phi(x)\phi(y)$.
Also, $\phi$ is injective (because $\phi(x)=\phi(y)$ implies $x\equiv y\pmod n$ and together with $x\equiv y\pmod m$ we get $x\equiv y\pmod {nm}$)
As $A$ and $U(n)$ are finite sets of the same cardinality, this implies that $\phi$ is a bijection
We conclude that $\phi$ is a magma isomorphism between $A$ and $U(n)$, and as $U(n)$ is a group, this meams that the magma $A$ is in fact a group and $\phi$ is a group isomorphism. $\square$
It may be worth noting that the neutral element of $A$ is not $m\cdot 1$, but rather the solution to $x\equiv 1\pmod n$, $x\equiv 0\pmod m$ that exists by virtue o fthe Chinese Remainder Theorem. (In your example, $25$)
The condition that $m$ and $n$ are coprime is necessary: If $d=\gcd(m,n)$, then the product of any two elements of $A$ is a multiple of $d^2$ (even after reduction modulo $mn=d^2\cdot \frac md\frac nd$), hence can never be $1\in A$.
There seems to be confusion here between the additive and multiplicative groups of $\mathbb Z/17$.
The additive group, i.e., the group whose operation is addition modulo 17, consists of all 17 congruence classes mod 17. So it is a group of prime order (namely 17), which implies that it is cyclic and every element, except for the identity 0, is a generator. (Check it: Randomly choose a non-zero congruence class, and repeatedly add; you'll get all 17 congruence classes. For example, if you start with 4, you get (in order) 4, 8, 12, 16, 3, 7, 11, 15, 2, 6, 10, 14, 1, 5, 9, 13, 0.)
The multiplicative group, i.e., the group whose operation is multiplication modulo 17, consists of the 16 non-zero congruence classes. It does not contain 0, because 0 has no multiplicative inverse (modulo 17). Since 16 isn't prime, you can't infer from general group theory that this group is cyclic, but it is a theorem that the multiplicative group of a finite field (like $\mathbb Z/17$) is cyclic. So the multiplicative group in your situation is a cyclic group of order 16.
In such a group, a cyclic group of non-prime order, some of the elements are generators (that's what it means to be cycic), but not all of them. For example, 6 happens to be a generator modulo 17. If you start with 6 and repeatedly multiply by it (reducing modulo 17 at every step), you get (in order) 6, 2, 12, 4, 7, 8, 14, 16, 11, 15, 5, 13, 10, 9, 3, 1. So 6 is a generator of the multiplicative group. On the other hand, if you do the same thing starting with 4, you get 4, 16, 13, 1 and then the numbers just repeat. So 4 is not a generator of your multiplicative group. Similarly, if you tart with 2, you get the computation in your question, showing that 2 is not a generator of the multiplicative group. (It generates a subgroup of order 8.)
The bottom line is that you did a computation in the multiplicative group, which has order 16, but then quoted a theorem that applies to groups of prime order like the additive group of order 17. If you keep straight which of the two groups you're talking about, and the fact that their orders are slightly different, your difficulties should vanish.
Best Answer
I think all you need has been pointed here. You don't need to write down the associated Cayley table for the presented set and its operation, but for this one it leads you to get the answer graphically:
We can see that the operation is a binary one. Can you find the identity element? What about the inverses ones?