For every $x$ the equation $x=yaya^{-1}$ has a solution. Indeed, replace $y$ by $za$. We get $x=z^2a^{-2}$ or $xa^2=z^2$. But in $SL_n(\mathbb{C})$ every element is a square, so $z$ can be found, hence $y=za^{-1}$ exists.
Update. In order to see that every complex non-singular matrix is a square, it is enough to consider a jordan block $J_n(a)$ ($a$'s on the diagonal, 1 on the next diagonal, the rest are 0). If we denote $b=\sqrt{a}$, then a square root of $J_n(a)$ is a triangular matrix, it has $b$ on the diagonal, $2/b$ on the next diagonal, $-\frac{1}{8b^3}$ on the second diagonal, $\frac{1}{16b^5}$ on the third diagonal, etc. The number on the diagonal # $i$ is equal to $c_i/(b^{2i-1})$ where $c_i$ can be defined by induction, it does not depend on $a$. Note that we need $a\ne 0$, so this does not work for singular matrices.