Are change of basis matrices characteristic of the matrix itself? Suppose we have a matrix $A$. If my understanding is correct, we could see the matrix in the canonical base of a linear transformation $f$ or a bilinear form $\phi$, for example.
So, given another base $B$, we know that
\begin{align} [\phi]_B &= T^t[\phi]_ET \end{align}
\begin{align} [f]_B &= T^{-1}[f]_ET \end{align}
where $T$ is the change of basis matrix from $B$ to $E$. 
As not necessarily $T^{-1}=T^t$, is the change of basis matrix characteristical of the matrix or does it really matters if we look at it whether as a linear transformation or a bilinear form? 
 A: The change of basis matrix is the same for either a linear transformation or a bilinear form. The matrix is constructed as to convert vectors whose coordinates are in $B$ back to the standard basis. This much is true in both cases.
As you indicate, however, the matrices $[\phi]_B$ or $[f]_B$ are different depending on whether you view a matrix as representing a linear transformation or a bilinear form. So in that sense it is not “intrinsic” to the matrix. This is because for a linear transformation you need to first transform into the new coordinates (apply $T$), apply the transformation, and then change back using $T^{-1}$. For a bilinear form, you need to convert both of the input vectors into the new coordinates by applying $T$ and $T^t$. 
There is an important example where the $[\phi]_B$ and $[f]_B$ coincide, namely when the bases are orthonormal. In which case the change of basis matrix is orthogonal $Q^t = Q^{-1}$. Outside of this case, though $[\phi]_B$ and $[f]_B$ will in general be different.
A: We are considering two kinds of relations between matrices here, as well as two interpretations of a matrix. It so happens that each kind of operation behaves well with respect to one of these interpretations, and less well with respect to the other.
To elaborate, the relation between $A$ and $T^{t}AT$, where $A$ is a square matrix with real entries, $T$ is an invertible square matrix of the same order as $A$, also with real entries, and $T^{t}$ denotes the transpose of the matrix $T$, is called congruence.
The relation between $A$ and $T^{-1}AT,$ with intepretations as above, and $T^{-1}$ the inverse of the matrix $T$, is similarity.
When we think of $A$ as representing a bilinear form, we see that the congruent matrix $T^{t}AT$ gives rise to the same bilinear form in a new set of coordinates: if $y=T^{-1}x,$ then $$y^{t}(T^{t}AT)y=x^{t}(T^{\phantom{.}}T^{-1})^{t}A(T^{\phantom{.}}T^{-1})x=x^{t}Ax.$$ On the other hand, this does not work when we consider a similar matrix (unless the similarity is through a matrix with $T^{-1}=T^{t}$), since if $y=T^{-1}x$, $$y^{t}(T^{-1}AT)y=x^{t}T^{-t}T^{-1}A(T^{\phantom{.}}T^{-1})x=x^{t}(TT^{t})^{-1}Ax\neq x^{t}Ax,$$ and if $y=T^{t}x,$ $$y^{t}(T^{-1}AT)y=x^{t}TT^{-1}ATT^{t}x=x^{t}A(TT^{t})x\neq x^{t}Ax.$$
When we think of $A$ as representing a linear transformation, we see that the similar matrix $T^{-1}AT$ gives rise to the same linear transformation in a new set of coordinates: if $y=T^{-1}x,$ then $$T^{-1}ATy=T^{-1}A(T^{\phantom{.}}T^{-1})x=T^{-1}Ax.$$ On the other hand, this does not work when we consider a congruent matrix (unless the congruence is through a matrix with $T^{-1}=T^{t}$), since if $y=T^{-1}x,$ $$T^{t}ATy=T^{t}A(T^{\phantom{.}}T^{-1})x=T^{t}Ax\neq T^{-1}Ax,$$ and if $y=T^{t}x,$ then $$T^{t}ATy=T^{t}A(T^{\phantom{.}}T^{t})x\neq T^{t}Ax.$$
