Is the metric induced by a norm ''unique''? Let $(X,\rVert{\cdot}\lVert)$ be a normed vector space. Clearly there is a canonical way to induce a metric (and so a topology) on $X$ by defining $d(x,y)=\rVert x-y\rVert$. 
Are there other metrics induced by  $\rVert{\cdot}\lVert$? If does exist another metric $d_1\neq d$ induced by $\rVert{\cdot}\lVert$, are $d$ and $d_1$ equivalent?
 A: Even if the question is loosely formulated, I would say that the answer is no. I'm thinking at the following examples: in the vector space $\mathbb{R}$, consider the usual absolute value $\lvert\cdot\rvert$, which indeed is a norm and induces the ordinary topology of the real line. We can define a metric on $\mathbb{R}$ in terms of $\lvert\cdot\rvert$ as follows: 
$$d_1(x, y)=\lvert \arctan(x)-\arctan(y)\rvert,$$
or as follows:
$$d_2(x, y)=\frac{\lvert x-y\rvert}{1+\lvert x-y\rvert}$$.
Both are "metrics induced by $\lvert \cdot\rvert$" in some vague sense, and neither of them is metrically equivalent to $\lvert \cdot\rvert$, because both make $\mathbb{R}$ into a bounded space. 
P.S. : I had not noticed that you consider "equivalent" two metrics which induce the same topology. Then $d_1$ and $d_2$ above do not qualify as counterexamples because they do induce the ordinary topology of the line. To obtain a counterexample in this case you can take the $\text{signum}$ function 
$$\text{signum}(s)=\begin{cases} 1& s>0 \\ 0 & s=0 \\ -1& s<0\end{cases}$$
and define 
$$d_3(x, y)=\text{signum}(\lvert x-y\rvert).$$
You get the discrete metric on $\mathbb{R}$, which obviously does not induce the same topology on $\mathbb{R}$ as $\lvert\cdot\rvert$.
