Theorem. Let $n\geq 3$. Then $5$ has order $2^{n-2}$ modulo $2^n$ and $2^{n}$ is the largest power of $2$ that divides $5^{2^{n-2}}-1$.
Proof. We proceed by induction. For $n=3$, $5\not\equiv 1\pmod{8}$, but $5^2=25\equiv 1\pmod{8}$. Note moreover that $2^3\Vert 5^2-1$.
Assume $5$ has order $2^{n-2}$ modulo $2^n$ and that $2^{n}\Vert (5^{2^{n-2}}-1)$. Then $5^{2^{n-2}}= 1+k2^{n}$ for some odd integer $k$, but $5^{2^{n-2}}\not\equiv 1\pmod{2^{n+1}}$.
Squaring we get
$$5^{2^{n-1}} = (1+k2^n)^2 = 1 + k2^{n+1}+2^{2n}\equiv 1\pmod{2^{n+1}},$$
so the order of $5$ modulo $2^{n+1}$ divides $2^{n-1}$, and is greater than $2^{n-2}$; thus, the order is precisely $2^{n-1}$. In addition, since $k$ is odd the largest power of $2$ that divides $5^{2^{n-1}}-1 = 2^{n+1}(k+2^{n-1})$ is $2^{n+1}$, completing the induction. $\Box$
Now note that all the powers of $5$ modulo $2^n$ are congruent to $1$ modulo $4$; The elements $-5$, $-(5^2),\ldots,-(5^{2^{n-2}})$ are all congruent to $3$ modulo $4$ and pairwise incongruent. In particular, there are two elements of order $2$ in $(\mathbb{Z}/2^n\mathbb{Z})^{\times}$, namely $5^{2^{n-3}}$ and $-5^{2^{n-3}}$. So $(\mathbb{Z}/2^n\mathbb{Z})^{\times}$ cannot be cyclic.
Since $(\mathbb{Z}/2^n)^{\times}$ has order $2^{n-1}$, is abelian, is not cyclic for $n\geq 3$, and has an element of order $2^{n-2}$, it follows that it is isomorphic to $C_{2^{n-2}}\times C_2$.
Best Answer
Addressing your question "I mean, is true that $A \ncong B \implies C \times A \ncong C \times B$?":
Yes, finite abelian groups are cancellable, see this post:
For groups $A,B,C$, if $A\times B$ and $A\times C$ are isomorphic do we have $B$ isomorphic to $C$?
So $A\times B\cong A\times C$ implies that $B\cong C$. So we would obtain $C_4\cong C_2\times C_2$, which is a contradiction, since one group is cyclic, the other not.