Here are two theorems:
Theorem 1: Let $\beta$ and $\beta'$ be two ordered bases for a finite-dimensional vector space V, and let $Q=[I_V]^\beta_{\beta '}$, then for any v $\in$ V, $[v]_\beta=Q [v]_{\beta'}$.
Theorem 2: Let T be a linear operator on a finite-dimensional vector space V, and let $\beta$ and $\beta '$ be ordered bases for V. Suppose Q is the change of coordinate matrix that changes $\beta '$-coordinates into $\beta$, then $[T]_\beta'=Q^{-1}[T]_\beta Q$
Proof of theorem 2: We have $Q[T]_{\gamma}=[I_v]^\beta_\gamma[T]_\gamma=[I_v T]^\beta_\gamma=[T I_v]^\beta_\gamma=[T]_\beta [I_v]^\beta_\gamma=[T]_\beta Q$.
Question: Do Q serve the same purposes in here?
Question 2: for the proof of theorem 2, why do we claim $Q[T]_\gamma$, is that from the result of theorem 1?
The key is to understand what exactly is going on with that bracket notation. For a transformation $T:V \to W$ and bases $\beta,\beta'$ of $V$ and $W$, $[T]^\beta_{\beta'}$ is defined to be the matrix for which $$ [T]^\beta_{\beta'}[v]_{\beta} = [T(v)]_{\beta'}. $$ Here, $[v]_{\beta}$ denotes the coordinate vector of $v$ relative to the basis $\beta$. On the other hand, the change of basis matrix $Q$ is defined so that we just get the input with respect to the new basis. In other words, we want $Q$ to satisfy $$ Q[v]_\beta = [v]_{\beta'}. $$ If we want to replace the $Q$ in this equation with $[T]^\beta_{\beta'}$ for some linear transformation $T$ on $V$, then the only transformation that makes sense is the identity map $I_V$ defined by $I_V(v) = v$. So, $Q = [I_V]^\beta_{\beta'}$.