What is the largest (finite) order of an element of $GL_{10}(\mathbb{Q})$? 
What is the largest (finite) order of an element of $GL_{10}(\mathbb{Q})$?

The following is my guess: Let $A \in GL_{10}(\mathbb Q)$ is of finite order $n$ (say). Then $A^n=I$, or every eigenvalue of $A$ is an $n$-th root of $1$. Thus an eigenvalue $\lambda$ of $A$ has degree  $\leq\phi(n)$ over $\mathbb Q$. Equality holds only when $\lambda$ is primitive $n$-th root of $1$. OTOH $\lambda$ also satisfies the characteristic polynomial of $A$, which is of degree $10$. Thus if the characteristic polynomial is irreducible (e.g., $A$ is companion matrix of $X^{10}+X^9+\cdots+X+1$) then degree of $\lambda=10$, then we must have $\phi(n)\leq10$. The maximum $n$ such that $\phi(n)=10$ is $15$.
If I am doing some mistake please correct me. Thanks.
 A: Briefly: the answer is $120$. One such matrix is
$$
A = 
\pmatrix{0&1\\& \ddots & \ddots \\
\\
\\
\\
\\
\\
\\
&&&&&&&\ddots & \ddots
\\
&&&&&&&&0&1
\\-1&-2 & -3 & -3 & -4 & -4 & -4 & -3 & -3 & -2}.
$$

Here is an approach to the problem.
The matrix $A$ will satisfy $A^n = I$ if and only if the minimal polynomial of $A$ divides $x^n - 1$. Notably, $x^n - 1$ is a product of cyclotomic polynomials (which are irreducible over $\Bbb Q$), so the minimal polynomial of $A$ must be a product of these. Because $A$ has size $10$, its minimal polynomial has degree at most $10$.
Let $\Phi_n$ denote the $n$th cyclotomic polynomial. The degree of $\Phi_n$ is $\phi(n)$ (where $\phi$ denotes the Euler totient function). For a product $p = \Phi_{n_1} \cdots \Phi_{n_k}$ with $n_1,\dots,n_k$ distinct, the smallest $n$ for which $x^n - 1$ is divisible by $p$ will be $n = \mathrm{lcm}(n_1,\dots,n_k)$.
Putting all that together, we can frame the problem as follows:

Maximize $\mathrm{lcm}(n_1,\dots,n_k)$ over all sets of distinct positive integers $n_1,\dots,n_k$ satisfying the constraint that $\phi(n_1) + \cdots + \phi(n_k) \leq 10$.

This maximal gcd will be equal to the largest possible finite order of an element of $GL_{10}(\Bbb Q)$.
Notably, there are lower bounds on $\phi$ that make it possible to obtain a solution by brute force. In particular, it is known that $\phi(n) \geq \sqrt{n/2}$. Thus, each $n_k$ must satisfy
$$
\sqrt{n_k/2} \leq \phi(n_k) \leq 10 \implies n_k \leq 200.
$$

After a systematic search I have found that the highest possible order is $120$. This maximum is attained with a matrix whose minimal polynomial is equal to
$$
\Phi_3\Phi_5\Phi_8 = (x^2 + x + 1)(x^4 + x^3 + x^2 + x + 1)(x^4 + 1).
$$
For instance, we can take $A$ to be the associated companion matrix.
Here's the Python code I used to compute this answer:
from math import gcd


def gcds(*args):
    if len(args) == 2:
        return gcd(*args)
    else:
        return gcds(gcd(*args[:2]),*args[2:])

def lcm(*args):
    if len(args) == 1:
        return args[0]
    if len(args) == 2:
        return args[0]*args[1]//gcd(*args)
    else:
        return lcm(lcm(*args[:2]),*args[2:])

def Ephi(n):
    s = 0
    for k in range(n):
        if gcd(k,n) == 1:
            s += 1
    return s

def get_list(deg, start = 1, M = 200):
    if deg == 0:
        return [[]]
    arr = []
    for k in range(start,M+1):
        phi = Ephi(k)
        if phi <= deg:
            new_deg = deg - phi
            new_M = 2*new_deg**2
            arr += [[k] + L for L in get_list(new_deg, start = k+1,  M = new_M)] # start=k allows repeats
    return arr

if __name__ == '__main__':
    
    n = 10 # set value of n (in GL(n,Q)) here

    M = 2*n**2
    arr = get_list(n, M=M)
    print('    number of non-derogatory elements with finite order (up to iso): ' + str(len(arr)))
    m = 0
    for L in arr: 
        x = lcm(*L)
        if x > m:
            m = x
            mL = L
    print('    max order: %d'%m) #prints maximal order
    s = ''
    for k in mL:
        s += 'Phi_%d '%k
    # minimal polynomial of element with maximal order as product of 
    # cyclotomic polynomials
    print('minimal polynomial of maximal element: ' + s)  

There are a few other candidates for the minimal polynomial. All together, the valid possibilies are
$$
\Phi_3\Phi_5\Phi_8,\\
\Phi_3\Phi_8\Phi_{10},\\
\Phi_5\Phi_6\Phi_8,\\
\Phi_6\Phi_8\Phi_{10}.
$$

Relevant paper: Finite Groups of Matrices Whose Entries Are Integers
