If $\alpha$ is a root, then $\sigma(\alpha)$ is a root There is the well-known proposition (for a specific reference: see section 14.1 of Dummit/Foote):
Proposition.  If $\sigma \in \text{Aut}(K/F)$ and $f(x) \in F[x]$ has $\alpha \in K$ as a root, then $f(x)$ also has $\sigma(\alpha)$ as a root.
This got me wondering about the following scenario:

Suppose there exists distinct $\sigma_1, \sigma_2 \in \text{Aut}(K/F)$ with the property that $\sigma_1(\alpha) = \sigma_2(\alpha)$, where $\alpha \in K$ is some root of $f(x)$.  Then does $f(x)$ have $\sigma_1(\alpha)$ as a root with multiplicity $2$? (i.e. one multiplicity "comes from" $\sigma_1$ and one multiplicity  "comes from" $\sigma_2$)

After some thought, I think this is false.  For example, consider a Galois extension $K/F$ and some irreducible polynomial $p(x) \in F[x]$ with $\alpha$ as a root.  If it so happens that there are distinct $\sigma_1, \sigma_2 \in \text{Aut}(K/F)$ with $\sigma_1(\alpha) = \sigma_2(\alpha)$, then it cannot be that $p(x)$ has $\sigma_1(\alpha)$ as a multiple root, since $K/F$ being Galois implies that $p(x)$ is separable.  
Is this reasoning correct?  Also, are there any easy counterexamples when $K/F$ is not Galois?  (If an example would require some exotic construction, then please don't waste your time writing it up...)
Thanks so much!

The reasoning is correct, but we can make a stronger observation. The proposition you mentioned can be strengthened as follows: If $\sigma \in \operatorname{Aut}(K/F)$, $f \in F[x]$, and $\alpha \in K$, then $\sigma(\alpha)$ has the same multiplicity as a root of $f$ as does $\alpha$.
Proof:
For any $p \in K[x]$, let $\sigma p$ be the polynomial obtained by applying $\sigma$ to the coefficients of $p$. This gives an automorphism of the ring $K[x]$.
Now, if $\alpha$ has multiplicity $n$, then $f(x) = (x - \alpha)^n g(x)$ for some $g \in K[x]$. Then, $f(x) = (\sigma f)(x) = (x - \sigma(\alpha))^n (\sigma g)(x)$. Thus $\sigma(\alpha)$ has multiplicity at least $n$, and applying the same reasoning with $\sigma^{-1}$ we see that the multiplicities are equal. QED
Therefore, in your proposed scenario, $\sigma_1(\alpha)$ cannot be a double root unless $\alpha$ itself is.
