Let $R = \Bbb Z/N\Bbb Z$. I'll closely follow this 2002 paper by V. P. Elizarov.
Lemma 4.2. If $a, b \in R$ and $\gcd(a,b)=d$ in $\Bbb Z$, then there exists a matrix $Q \in GL_2(R)$ s.t. $(a,b)Q = (d,0)$.
Proof. If $a = 0$ or $b = 0$, take $Q$ to be the identity or an antidiagonal matrix with $1$'s. Suppose $a, b \not = 0$. Over $\Bbb Z$, express $d = a u_1 + b v_1$ by Bezout's identity. If $a = df$ and $b = dg$, then $1 = f u_1 + g v_1$ in $\Bbb Z$.
Furthermore, denote $u = u_1 \mod N$ and $v = v_1 \mod N$ ($u,v \in R$); then we have $d = au+bv$ in $R$, $1=fu+gv$ in $R$. Take $Q = (\begin{smallmatrix}u&g\\v&-f\end{smallmatrix})$ (it seems that the second column in his paper was upside down). Then $AQ = (au + bv, ag-bf) = (1,0)$; the determinant is $-uf-vg=-1$ which means $Q$ is invertible.
Statement 4.3. If $\gcd(a_1, \ldots, a_n) = d$, then there exists $V \in GL_n(R)$ s.t. $AV = (d, 0, \ldots, 0)$, where $A = (a_1,\ldots,a_n)$.
Proof. Start from $(a_{n-1}, a_n)$ and successively eliminate the nonzero elements using Lemma 4.2 (going backwards). Then take the product of these matrices, padded with blocks of identity transformations.
Special case of Statement 4.4. The set of solutions of $A (x_1, \ldots, x_n)^T = 0$ over $R$, where $\gcd(\{a_i\}, n) = 1$, is $V (0, *, \ldots, *)^T$ (the asterisks are arbitrary elements of $R$).
The proof is said to be "a direct check".
Denote $S = (0, *, \ldots, *)^T$. Then, for $A$ and $B$, we can find the corresponding matrices $V_A$ and $V_B$ and ask whether $V_A S = V_B S$, or equivalently, $V_B^{-1}V_A S = S$. Both sides have equal cardinalities, so we only need to check that $V_B^{-1} V_A e_k \in S$ for $k = 2, \ldots, n$ ($\{e_k\}$ is a "standard basis" of $S$, i.e. vectors of all zeros except one $1$).
Best Answer
Not true. If $(x_i)$ is a basis so is $(2x_i)$. See what happens in this case. However if the bases are orthonormal bases then the conclusion is true and both sums are equal to $\|x\|$.