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Here is an explicit example to show that the answer to the first question os "no" in general:

Let $m$ be an integer greater than $2$. Let $K = \left(\begin{array}{clcr} 1 & m\\0 & 1\end{array}\right)$ and let $B = \left(\begin{array}{clcr} 0 & -1\\1 & m\end{array}\right)$. Then $B^{T}KB = K$, but $B$ is not a matrix of finite order, ( its eigenvalues are real, but neither has absolute value $1$). The fact that $KBK^{-1} = (B^{T})^{-1}$ if your equality holds is what led to this example, for then, in particular, $B$ and $B^{-1}$ must have the same eigenvalues.

Later edit: Note that when $K = I_{n}$, then $|C^{\prime}(K)| = 2^{n}n!$, which is probably the maximal possible order of such a group in the case that it is indeed finite (at least for large enough $n$). It is easy to prove (an argument of Blichfeldt), that every finite subgroup of ${\rm GL}(n,\mathbb{Z})$ hos order a divisor of $(2n)!$. The above group of order $2^{n}n!$ is sometimes known as the group of "signed permutation matrices".

Here is an explicit example to show that the answer to the first question is "no" in general:

Let $m$ be an integer greater than $2$. Let $K = \left(\begin{array}{clcr} 1 & m\\0 & 1\end{array}\right)$ and let $B = \left(\begin{array}{clcr} 0 & -1\\1 & m\end{array}\right)$. Then $B^{T}KB = K$, but $B$ is not a matrix of finite order, ( its eigenvalues are real, but neither has absolute value $1$). The fact that $KBK^{-1} = (B^{T})^{-1}$ if your equality holds is what led to this example, for then, in particular, $B$ and $B^{-1}$ must have the same eigenvalues.

Later edit: Note that when $K = I_{n}$, then $|C^{\prime}(K)| = 2^{n}n!$, which is probably the maximal possible order of such a group in the case that it is indeed finite (at least for large enough $n$). It is easy to prove (an argument of Blichfeldt), that every finite subgroup of ${\rm GL}(n,\mathbb{Z})$ has order a divisor of $(2n)!$. The above group of order $2^{n}n!$ is sometimes known as the group of "signed permutation matrices".

Here is an explicit example to show that the answer to the first question os "no" in general:

Let $m$ be an integer greater than $2$. Let $K = \left(\begin{array}{clcr} 1 & m\\0 & 1\end{array}\right)$ and let $B = \left(\begin{array}{clcr} 0 & -1\\1 & m\end{array}\right)$. Then $B^{T}KB = K$, but $B$ is not a matrix of finite order, ( its eigenvalues are real, but neither has absolute value $1$). The fact that $KBK^{-1} = (B^{T})^{-1}$ if your equality holds is what led to this example, for then, in particular, $B$ and $B^{-1}$ must have the same eigenvalues.

Later edit: Note that when $K = I_{n}$, then $|C^{\prime}(K)| = 2^{n}n!$, which is probably the maximal possible order of such a group in the case that it is indeed finite (at least for large enough $n$). It is easy to prove (an argument of Blichfeldt), that every finite subgroup of ${\rm GL}(n,\mathbb{Z})$ hos order a divisor of $(2n)!$. The above group of order $2^{n}n!$ is sometimes known as the group os "signed permutation matrices".

Here is an explicit example to show that the answer to the first question os "no" in general:

Let $m$ be an integer greater than $2$. Let $K = \left(\begin{array}{clcr} 1 & m\\0 & 1\end{array}\right)$ and let $B = \left(\begin{array}{clcr} 0 & -1\\1 & m\end{array}\right)$. Then $B^{T}KB = K$, but $B$ is not a matrix of finite order, ( its eigenvalues are real, but neither has absolute value $1$). The fact that $KBK^{-1} = (B^{T})^{-1}$ if your equality holds is what led to this example, for then, in particular, $B$ and $B^{-1}$ must have the same eigenvalues.

Later edit: Note that when $K = I_{n}$, then $|C^{\prime}(K)| = 2^{n}n!$, which is probably the maximal possible order of such a group in the case that it is indeed finite (at least for large enough $n$). It is easy to prove (an argument of Blichfeldt), that every finite subgroup of ${\rm GL}(n,\mathbb{Z})$ hos order a divisor of $(2n)!$. The above group of order $2^{n}n!$ is sometimes known as the group of "signed permutation matrices".

Here is an explicit example to show that the answer to the first question os "no" in general:

Let $m$ be an integer greater than $2$. Let $K = \left(\begin{array}{clcr} 1 & m\\0 & 1\end{array}\right)$ and let $B = \left(\begin{array}{clcr} 0 & -1\\1 & m\end{array}\right)$. Then $B^{T}KB = K$, but $B$ is not a matrix of finite order, ( its eigenvalues are real, but neither has absolute value $1$). The fact that $KBK^{-1} = (B^{T})^{-1}$ if your equality holds is what led to this example, for then, in particular, $B$ and $B^{-1}$ must have the same eigenvalues.

Here is an explicit example to show that the answer to the first question os "no" in general:

Let $m$ be an integer greater than $2$. Let $K = \left(\begin{array}{clcr} 1 & m\\0 & 1\end{array}\right)$ and let $B = \left(\begin{array}{clcr} 0 & -1\\1 & m\end{array}\right)$. Then $B^{T}KB = K$, but $B$ is not a matrix of finite order, ( its eigenvalues are real, but neither has absolute value $1$). The fact that $KBK^{-1} = (B^{T})^{-1}$ if your equality holds is what led to this example, for then, in particular, $B$ and $B^{-1}$ must have the same eigenvalues.

Later edit: Note that when $K = I_{n}$, then $|C^{\prime}(K)| = 2^{n}n!$, which is probably the maximal possible order of such a group in the case that it is indeed finite (at least for large enough $n$). It is easy to prove (an argument of Blichfeldt), that every finite subgroup of ${\rm GL}(n,\mathbb{Z})$ hos order a divisor of $(2n)!$. The above group of order $2^{n}n!$ is sometimes known as the group os "signed permutation matrices".