Is there a $B\in O(n-1)$ such that $A \cdot \pmatrix{\pm1& 0 \\\ 0& B} = A'$?

Question

I was wondering if the following is true:

Suppose we have matrices $A, A' \in O(n)$ such that $A e_1 = \pm A' e_1$. Is there a $B\in O(n-1)$ such that $$A \cdot \pmatrix{\pm1& 0 \\\ 0& B} = A'$$ ?

Does such a $B$ exist? How could we construct it?

Answer 1

Put $C := (A')^{-1}A$. Since $A,\ A' \in O(n)$, we have $C \in O(n)$. This is equivalent to the statement that the columns of $C$ are orthonormal. Moreover $Ae_1 = \pm A'e_1$ implies $Ce_1 = (A')^{-1}Ae_1 = \pm e_1$. This says that the first column of $C$ has $\pm1$ at the top and $0$ elsewhere. Also, the other columns of $C$ must be orthogonal to the first column, so they must have $0$ at the top. This in turn implies that the columns of the bottom right $(n-1)\times (n-1)$-matrix of $C$ are orthogonal to each other and every one of them has norm $1$. This means that the bottom right $(n-1)\times (n-1)$-matrix of $C$ is in $O(n-1)$. All in all we see that $C$ has the desired form.