Solution for $x^9 + x^6 + x^4 - x^2 + 1=0$

You are correct. Not all roots of the same polynomial are necessarily "equal".

In algebra, we encode this information in so-called Galois group of the field extension $Gal (F:\mathbb Q) $, where $F $ is the smallest field containing all of $\mathbb Q $ and, in addition, all of the roots. It turns out that the elements of that group are some (but not always all!) permutations of the roots.

In your case, the roots of $x^5-x+1$ can only swap with the other roots of $x^5-x+1$, and the roots of $x^4+x+1$ can only swap with the other roots of $x^4+x+1$. In essence, your Galois group is definitely not the whole $S_9$ but is a direct product of some subgroup of $S_5$ and some subgroup of $S_4$.

There are other polynomials of degree 9, however, for which the Galois group is the whole of $S_9$ (all permutations) and for which all the roots are "equal".

Since you have mentioned Abel-Ruffini theorem and user8734617 has mentioned Galois group, I will also mention some Galois theory. As you have noticed, your degree $9$ polynomial can be factored into a product of degree $4$ and degree $5$ polynomial. Indeed, the degree $4$ polynomial is solvable by radicals using Ferrari's method.

However, it is not the case that any degree $5$ polynomial is not solvable by radicals. In fact, Galois's solvability theorem extends Abel-Ruffini theorem by telling us that in a field $F$ with characteristic $0$, any non-constant polynomial $f$ over $F$ is solvable by radicals if and only if its Galois group over $F$, $\operatorname{Gal}(f / F)$, is solvable. Here we are interested in the case $F = \mathbb{Q}$, the rational numbers.

So to prove $x^5 - x + 1$ is not solvable by radicals, we need to show $\operatorname{Gal}(x^5 - x + 1 / \mathbb{Q})$ is not solvable. Our strategy is to show $\operatorname{Gal}(x^5 - x + 1 / \mathbb{Q}) = S_5$, the symmetric group on $5$ letters. Since $\operatorname{Gal}(x^5 - x + 1 / \mathbb{Q})$ is always contained in $S_5$, it suffices to show $S_5$ is contained in $\operatorname{Gal}(x^5 - x + 1 / \mathbb{Q})$.

From this post, $x^p - x + a$ where $a \ne 0$ is irreducible$\bmod{p}$. So we have $x^5 - x + 1$ is irreducible$\bmod{5}$ by letting $p = 5$ and $a = 1$.

Besides, $x^5 - x + 1 = (x^2 + x + 1) (x^3 + x^2 + 1) \bmod{2}$ where both $x^2 + x + 1$ and $x^3 + x^2 + 1$ are irreducible$\bmod{2}$ since they have no roots$\bmod{2}$.

By Dedekind's theorem, $\operatorname{Gal}(x^5 - x + 1 / \mathbb{Q})$ contains permutation with cycle type $(a\ b)(c\ d\ e)$ and a $5$-cycle. Now $((a\ b)(c\ d\ e))^3 = (a\ b)^3(c\ d\ e)^3 = (a\ b)$. So $\operatorname{Gal}(x^5 - x + 1 / \mathbb{Q})$ contains a transposition and $5$-cycle, which means $S_5$ is contained in $\operatorname{Gal}(x^5 - x + 1 / \mathbb{Q})$.

Finally, since $S_5$ is not solvable, we conclude $x^5 - x + 1$ is not solvable by radicals!