Endomorphism and Matrices In the context of linear maps I need help to understand the notation used in this exercise and the general method to solve it.

In the Euclidean space $\mathbb R^3$ with the standard inner product, consider the endomorphism $f : \mathbb R^3 \rightarrow \mathbb R^3$ defined by
$$f(1,0,1) = (1,4,1), \,\, f(1,0,-1) = (-3,0,3), \,\,f(0,1,0)=(2,-1,2).$$
Find the associated $f^*$.

Since I'm translating directly from greek notes I'm not sure if the word &quot;associated&quot; is correct in this context. Nowhere in my notes does the author gives the definition of $f^*$. This is the major problem that I have and therefore, this is my first question:
$(Q1)$ What is the precise definition of $f^*$? Specifically, what does the exercise ask us to find? Also, if the word &quot;associated&quot; is not correct, what is the correct one?
The author gives the following quick solution to the exercise:
Solution. The matrix of $f$ in the standard basis $B$ of $\mathbb R^3$ is
$$A = M_B^B(f) = \begin{pmatrix}-1 &amp; 2 &amp; 2\\2 &amp; -1 &amp; 2\\2 &amp; 2 &amp;-1\end{pmatrix}.$$
Since the matrix $A$ is symmetric it follows that $f = f^*$.
Therefore, the associated $f^* : \mathbb R^3 \rightarrow \mathbb R^3$ of $f$ is exactly $f$.

$(Q2)$ At the moment the above solution doesn't make any sense to me. I can see that the matrix $A$ is symmetric but I have no idea how to find it, what it actually represents, how it is connected to the exercise and what to conclude from it.
I'm looking for answers to $(Q1)$ and $(Q2)$ which will allow me to understand the exercise and solution.
As a linear transformation, $f$ can be represented through a matrix, whose columns are the images of the standard basis vectors. By linearity \begin{align}f(1,0,0)&amp;=f\left(\frac12(1,0,1)+\frac12(1,0,-1)\right)=^{\text{linearity of $f$}}\\[0.2cm]&amp;=\frac12f(1,0,1)+\frac12f(1,0,-1)=\frac12(1,4,1)+\frac12(-3,0,3)=(-1,2,2)\end{align} and similarly for $(0,1,0)$ and $(0,0,1)$. That is how he derived matrix $A$. Then $f^*$ is just the conjugate of $f$, the word "associated" seems unnecessary. Given the matrix $A$, the conjugate is found through the relation $$A^*=\overline{A}^T=A^T$$ since $A$ is real. For any $x,y \in \mathbb R^3$ it satisfies the relation $\langle Ax,y\rangle=\langle x,A^*y\rangle$ where $\langle \cdot, \cdot \rangle$ stands for the inner-product in $\mathbb R^3$, but most likely you do not need that in this question.
Basically this explains how to find a matrix representation of a function 
f(1,0,1)=(1,4,1),f(1,0,−1)=(−3,0,3),f(0,1,0)=(2,−1,2).The matrix representation is a transformation matrix which depends upon the basis vectors chosen. So for simplicity we choose the standard basis (1,0,0), (0,1,0) and (0,0,1).Then express the three given functions ( the function vector as taken together) in linear combinations of the basis vectors. So we find the functional values
 f(1,0,1)=(1,4,1),f(1,0,−1)=(−3,0,3),f(0,1,0)=(2,−1,2).
f(1,0,1)=(1,4,1),f(1,0,−1)=(−3,0,3),f(0,1,0)=(2,−1,2).
This expressed as a matrix becomes A = M(f) as above. The matrix is symmetric because of linearity of the function f