If two $n\times n$ matrices $A$ and $B$ are similar can we also conclude that there exists an invertible matrix $C$ of determinant = 1 such that $CAC^{-1} = B$ ?
If not is there a way to detect if two similar matrices are also similar via conjugation in $SL_n$ ?
If $n$ is odd or if you're working over $\mathbb{C}$, then yes.
Let $A$ and $B$ be similar $n \times n$ matrices and let $S$ be an invertible matrix such that $S^{-1}AS = B$. Letting $k = \mathrm{det}(S)$ and $C = \frac{1}{\sqrt[n]{k}}S$, we see that $\mathrm{det}(C) = 1$ and $C^{-1}AC = B$.
If $n$ is even and you're working over $\mathbb{R}$, then no.
For example $$A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \text{ and } B = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$$ are similar, but any invertible matrix $S$ such that $S^{-1}AS = B$ must have $\mathrm{det}(S) < 0$. To see this, note that if $S = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and $S^{-1}AS=B$, then writing out $AS=SB$ gives $a=d=0$ and $b=c$, so that $\mathrm{det}(S) = -b^2 < 0$.