Proof of orthogonal transformation Let $\alpha_1$,$\alpha_2$,...,$\alpha_s$ and $\beta_1$,$\beta_2$,..,$\beta_s$ be two families of vectors in $n$ dimensional Euclidean space $V$. Prove that the necessary and sufficient condition for an orthogonal transformation $\mathcal{A}$ to exist, such that
              \begin{equation}
               \mathcal{A}\alpha_i=\beta_i\qquad(i=1,2,...,s),
          \end{equation}
is that 
              \begin{equation}
               (\alpha_i,\alpha_j)=(\beta_i,\beta_j)\qquad(i,j=1,2,...,s).
          \end{equation}
The proof of necessity is easy. According to the definition of orthogonal transformation, for any inner product, we have 
               \begin{equation}
               (\alpha_i,\alpha_j)=(\mathcal{A}\alpha_i,\mathcal{A}\alpha_j)=(\beta_i,\beta_j)\qquad(i,j=1,2,...,s).
          \end{equation}
 A: The hypothesis implies that for any scalars $x_1,x_2,\ldots,x_s,y_1,y_2,\ldots,y_s$, we have
$$
\sum_{i=1}^s \sum_{j=1}^s x_iy_j(\alpha_i,\alpha_j)=
\sum_{i=1}^s \sum_{j=1}^s x_iy_j(\beta_i,\beta_j), \tag{1}
$$
or in other words,
$$
\bigg(\sum_{i=1}^s x_i\alpha_i , \sum_{j=1}^s y_j\alpha_j \bigg)=
\bigg(\sum_{i=1}^s x_i\beta_i , \sum_{j=1}^s y_j\beta_j \bigg) \label{2}\tag{2}
$$
If we take $y_k=x_k$ in \eqref{2}, we deduce
$$
\bigg|\bigg| \sum_{k=1}^s x_k\alpha_k \bigg|\bigg|=
\bigg|\bigg| \sum_{k=1}^s x_k\beta_k \bigg|\bigg| \tag{3}
$$
and hence
$$
\sum_{k=1}^s x_k\alpha_k = 0 \Leftrightarrow \sum_{k=1}^s x_k\beta_k = 0 \label{4}\tag{4}
$$
Since $\sum_{k=1}^s x_k\alpha_k=\sum_{k=1}^s y_k\alpha_k$ is equivalent to
$\sum_{k=1}^s (x_k-y_k)\alpha_k=0$, we deduce that
$$
\sum_{k=1}^s x_k\alpha_k = \sum_{k=1}^s y_k\alpha_k \Leftrightarrow \sum_{k=1}^s x_k\beta_k = \sum_{k=1}^s y_k\beta_k \label{5}\tag{5}
$$
Thus \eqref{5} tells us that there is a well-defined transformation $\cal A$ that sends every $\sum_{k=1}^s x_k\alpha_k$ to the corresponding $\sum_{k=1}^s x_k\beta_k $. And $\cal A$ is linear by construction.
If we put $x=\sum_{i=1}^s x_i\alpha_i$ and $y=\sum_{j=1}^s y_j\alpha_j$, we can now rewrite \eqref{2} as
$$
(x,y)=({\cal A}(x),{\cal A}(y)) \tag{6}
$$
So that $\cal A$ is an orthogonal transformation as wished.
A: The way you phrased it, the condition is only necessary for $A$ to be orthogonal, it is not sufficient. As a counterexample, consider the linear map $A\colon \mathbb R^3\to \mathbb R^3$ defined by
$$
A(x, y, z)=(x, y, 0),$$
which is not orthogonal. However, the family $\alpha_1=(1, 0, 0), \alpha_2=(0, 1, 0)$ is mapped into itself, thus, with your notation, $\beta_1=\alpha_1, \beta_2=\alpha_2$ and obviously $(\alpha_i, \alpha_j)=(\beta_i, \beta_j)$ for all $i, j=1, 2$.
To fix the statement, you need to require that the family $\alpha_1, \ldots, \alpha_n$ is a basis of $V$. If this is the case, then every vector $x, y\in V$ can be written in a unique way as
$$
x=\sum_{j=1}^n x_j \alpha_j, \quad y=\sum_{j=1}^n y_k \alpha_k.$$
Using the assumption that $(A\alpha_j, A\alpha_k)=(\alpha_j, \alpha_k)$ we obtain
$$
\begin{split} 
(Ax, Ay)&= \sum\sum x_j y_k (A\alpha_j, A\alpha_k)\\
&= \sum\sum x_j y_k (\alpha_j, \alpha_k)\\
&=(\sum x_j\alpha_j, \sum y_k \alpha_k)=(x, y).\end{split}$$
This proves that $A$ is orthogonal.
A: Define 
$A_{(n\times s)}\equiv(\alpha_1, \alpha_2, ..., \alpha_s)$
$B_{(n\times s)}\equiv(\beta_1, \beta_2, ..., \beta_s)$
Write $A$ into $A = E_A \cdot S_A = O_A \cdot I_{k_A} \cdot S_A$
Here 


*

*$E_A$ is the orthnomal base of $A$ after the Gram-Schmidt process

*$S_A$ is the matrix represent the Gram-Schmidt process operations for $A$.

*$I_{k_A} = \text{diag}\{1,1,...,1,0,...,0\}$ is a diagonal matrix with the first $k_A$ (the rank of A) elements being $1$ and the rest being $0$.

*$O_A$ is the orthogonal matrix that transform $I_{k_A}$ to $E_A$.


Because $(\alpha_i,\alpha_j) = (\beta_i, \beta_j)$, 


*

*$A$ and $B$ share the same rank, i.e. $k_A = k_B$

*$A$ and $B$ share the same Gram-Schmidt process operations. i.e. $S_A = S_B$
So now if we define the orthogonal matrix $\mathcal{A} = O_B \cdot O_A^T$ 
Then we have
$\mathcal{A}\cdot A = O_B \cdot O_A^T \cdot O_A \cdot I_k \cdot S  = O_B \cdot I_k \cdot S = B$
which just means $\mathcal{A}\cdot\alpha_i = \beta_i$,  $\forall i\in(1,2,...,s)$
A: The credit of this answer completely goes to Ewan Delanoy. I found that answer difficult to understand at first. Therefore, this is a simplified version
of that answer.
Let $V_{\alpha}$ be the space spanned by vectors $\alpha_{1}, \alpha_{2}, \dotsc, \alpha_{s}$. Let $V_{\beta}$ be the space spanned by vectors $\beta_{1}, \beta_{2}, \dotsc, \beta_{s}$. It is given that $(\alpha_{i},\alpha_{j}) = (\beta_{i},\beta_{j})$ for all $i,j \in \{1,\dotsc,s\}$. We want to find an orthogonal transformation $\mathcal{A}$ from $V_{\alpha}$ to $V_{\beta}$.
The Transformation: Any vector $x$ in $V_{\alpha}$ can be expressed in the following form $x = \sum_{i = 1}^{s} a_{i} \cdot \alpha_{i}$ for some scalars $a_{i} \in \mathbb{R}$. We define the transformation $\mathcal{A}$ as follows:
$$\mathcal{A}(x) = \sum_{i = 1}^{s} a_{i} \cdot \beta_{i}$$.
It is easy to see that $\mathcal{A}(x) \in V_{\beta}$ since it is a linear combination of vectors $\beta_{1}, \dotsc, \beta_{s}$. Moreover, $\alpha_{j}$ is mapped to $\beta_{j}$ since we can set $a_{j} = 1$ and $a_{i} = 0$ for all $i \in \{1,\dotsc,s\} \setminus \{j\}$.
To show that $\mathcal{A}$ is orthogonal, we need to prove that it is a linear transformation and it preserves the inner product. The proof goes as follows:

*

*$\mathcal{A}$ is Linear Transformation: We need to show that for any two vectors $x,y \in V_{\alpha}$,  $\mathcal{A}(x+y) = \mathcal{A}(x) + \mathcal{A}(y)$ and $\mathcal{A}(c \cdot x) = c \cdot \mathcal{A}(x)$ for any constant $c \in \mathbb{R}$. Using the above definition of $\mathcal{A}$ it is easy to prove both these properties. Therefore, $\mathcal{A}$ is a linear transformation.


*$\mathcal{A}$ preserves the inner product: We need to show that for any two vectors $x,y \in V_{\alpha}$, $(x,y) = (\mathcal{A}(x),\mathcal{A}(y))$. Let $x = \sum_{i = 1}^{s} a_{i} \cdot \alpha_{i}$ and $y = \sum_{i = 1}^{s} b_{i} \cdot \alpha_{i}$. The proof goes as follows:
\begin{align*}
(x,y) &= \left(\sum_{i = 1}^{s} a_{i} \cdot \alpha_{i} \right)^{T} \cdot \left(\sum_{i = 1}^{s} b_{i} \cdot \alpha_{i} \right) \\
&= \sum_{i = 1}^{s} \sum_{j = 1}^{s} a_{i} \cdot b_{i} \cdot (\alpha_{i},\alpha_{j}) \\
&= \sum_{i = 1}^{s} \sum_{j = 1}^{s} a_{i} \cdot b_{i} \cdot (\beta_{i},\beta_{j}) \quad \textrm{$\because$ it is given that $(\alpha_{i},\alpha_{j}) = (\beta_{i},\beta_{j})$}\\
&= \left(\sum_{i = 1}^{s} a_{i} \cdot \beta_{i} \right)^{T} \cdot \left(\sum_{i = 1}^{s} b_{i} \cdot \beta_{i} \right) \\
&= (\mathcal{A}(x))^{T} \cdot (\mathcal{A}(y)) \\
&= (\mathcal{A}(x),\mathcal{A}(y))
\end{align*}
This proves that $\mathcal{A}$ is an orthogonal transformation.

Note that any vector $x \in \mathcal{A}$ can be expressed in multiple forms. However, it will still get mapped to the same $\mathcal{A}(x)$. In other words, the transformation $\mathcal{A}$ is bijective. To prove that $\mathcal{A}$ is one-to-one; suppose a vector $x \in \mathcal{A}(x)$ has two forms say $\sum_{i = 1}^{s}a_{i} \cdot \alpha_{i}$ and $\sum_{i = 1}^{s}b_{i} \cdot \alpha_{i}$; then both these forms will map to the same $\mathcal{A}(x)$. For the sake of contradiction, suppose these two maps to vectors say $y = \sum_{i = 1}^{s}a_{i} \cdot \beta_{i}$ and $z = \sum_{i = 1}^{s}b_{i} \cdot \beta_{i}$; and $y\neq z$. Then $\| y-z\| \neq 0$. It contradicts the fact that $\mathcal{A}$ preserves the pairwise distances. In the same way, we can show that function is onto. This proves that $\mathcal{A}$ is indeed bijective.
