If $A$ is a square matrix such that $A^{27}=A^{64}=I$ then $A=I$

Firstly, since $A A^{26} = A^{26}A=I$, then $A$ is an invertible matrix.

Use the fact that $\gcd(27,64)=1$: hence there exist some $a,b \in \Bbb{Z}$ such that $1=27a+64b$. Now, compute $$A=A^1=A^{27a+64b}=(A^{27})^a(A^{64})^b=I^aI^b=I$$

\begin{align*} & I = A^{64} = A^{2(27) + 10} = (A^{27})^2A^{10} = A^{10}\\ \implies & I = A^{27} = A^{2(10) + 7} = (A^{10})^2A^7 = A^7\\ \implies & I = A^{10} = A^{1(7)+3} = (A^7)^1A^3 = A^3\\ \implies & I = A^7 = A^{2(3) + 1} = (A^3)^2A = A. \end{align*}

This is nothing more than the Euclidean algorithm applied to the exponents. The same procedure can be used to show that if $A^p = I$ and $A^q = I$ with $p$ and $q$ coprime, then $A = I$.

Hint: if matrix $A$ has eigenpair $(\lambda, v )$, then $A^{27}$ and $A^{64}$ have eigenpairs $(\lambda^{27}, v)$ and $(\lambda^{64}, v)$. At the same time $A^{27} v = A^{64} v = 1 \cdot v$. Could you proceed from here?