Let $F$ be a field and let $f(x) \in F[x]$ be a polynomial. Suppose that there is a field extension $K/F$ such that $f(x)$ factors as $(x-b_1)(x-b_2)\cdots (x-b_n)$ in $K$. Among the roots $b_i$, there might be some repeated roots (e.g. if we take $F = \mathbb{Q}$ and $f(x) = (x^2+x+1)^2 = (x - \omega)^2 (x - \omega^2)^2$).
We know that any $F$-automorphism of $K$ must permute the multiset $\{b_1, b_2, \cdots, b_n\}$. Suppose $\alpha, \beta$ are two of the roots appearing in the multiset with multiplicity $n_a, n_b$ respectively.
Claim: It's not possible for an automorphism $\sigma$ to send $\alpha$ to $\beta$ while satisfying $n_a \not= n_b$.
I think this follows from the following facts:
Fact 1. If $\sigma$ sends $\alpha$ to $\beta$ then $\alpha, \beta$ have the same minimal polynomial.
Fact 2. If $\alpha$ is a root of $f$ with multiplicity $n_a$, then all the conjugates of $\alpha$ are also roots of $f$, each with multiplicity $n_a$
Your thought is right.
Suppose there exists $\sigma\in\mathrm{Gal}(K/F)$ such that $\sigma(\alpha)=\beta$, then $\alpha,\beta$ obviously have the same minimal polynomial, which is denoted by $g(x)\in F[x]$.
Write $f(x)=g(x)^{N}h(x)$, where $h(x)$ is not divided by $g(x)$. Since factors $x-\alpha$ and $x-\beta$ only appear in $g(x)^{N}$, it suffices to prove that in $g(x)$ the multiplicities of $\alpha$ and $\beta$ are the same. If $char F=0$, this is obvious since $g$ is separable; Otherwise, $g(x)=(\widetilde{g}(x))^{p^{m}}$ for some $m$ and separable irreducible polynomial $\widetilde{g}$. (We can also just take the splitting field of $g(x)$ and consider the permutation of all its roots)
So $n_{\alpha}=n_{\beta}$ if $\alpha$ conjugates with $\beta$.