If $\rho_1$ and $\rho_2$ are two faithful positive linear functionals on a unital $C^*$-algebra $A$. Then $\rho_i(x)=\langle x\omega_i,\omega_i \rangle$, where $\omega_i=1+N_{\rho_i}$, $N_{\rho_i}=\{x\in A: \rho_i(x^*x)=0\}(i=1, 2).$
Since each $\rho_i$ is faithful, we have $N_{\rho_i}=0$. Then each cycic vector $\omega_i=1(i=1, 2),$ $\rho_1$ and $\rho_2$ have the same GNS construction.
Does the above conclusion correct?
On
Martin's example says it all but here is another, more pedestrian counter-example.
Let $A=C([0, 1])$ and let $D_1$ and $D_2$ be two disjoint, dense subsets of $[0, 1]$. For example, take $$ D_1=\mathbb Q\cap [0, 1], \quad \text{and} \quad D_2=(\mathbb Q+\pi )\cap [0, 1]. $$
Fix an enumaration $\{d_n^i\}_{n\in {\bf N}}$ for each $D_i$, and define a state $\rho _i$ on $C([0, 1])$ by $$ \rho _i(f) = \sum_{n\in {\bf N}} 2^{-n}f(d^i_n), \quad\forall f\in C([0, 1]). $$
It is not hard to see that the GNS representation $\pi _i$ associated to $\rho _i$ is equivalent to the representation of $C([0, 1])$ on $\ell ^2$, where each $f$ in $C([0, 1])$ is mapped to the diagonal operator with diagonal entries $f(d^i_n)$.
This implies that $\pi _i$ is faithful, so one might wonder whether or not $\pi _1$ and $\pi _2$ are equivalent. To see that they are not, it is enough to observe that, for any given $f$, the point spectrum (eigenvalues) of the operator $\pi _i(f)$ is $$ \sigma _p(\pi _i(f))= \{f(d^i_n): n\in {\bf N}\}, $$ and if $f$ is any injective function, this imlies that $$ \sigma _p(\pi _1(f))\neq \sigma _p(\pi _2(f)). $$
This shows that the operators $\pi _1(f)$ and $\pi _2(f)$ are not conjugate to each other, and hence $\pi _1$ cannot be equivalent to $\pi _2$.
No, even if $\rho_1$ and $\rho_2$ are states. For instance take $A=\text{UHF}(2^\infty)$ and take $\rho_1$ to be the trace and $\rho_2$ to be the Powers' state $\phi_\lambda$. Then $\pi(A)''$ is the hyperfinite II$_1$-factor, while $\rho_2(A)$ is a type III$_\lambda$ factor.