Lifting an isomorphism of quotients to an automorphism of free group

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Suppose that $F$ is a free group on $2$ generators and let $K,H \unlhd F$ be such that $F/H$ and $F/K$ are isomorphic.

If $\phi \colon F/H \to F/K$ is an isomorphism, can it be lifted to an automorphism $\tilde{\phi} \in \mathop{Aut}(F)$ such that $\tilde{\phi}(H) = K$?

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Not necessarily. For example, let $F=\langle a,b \rangle$ and let $H=K$ be the normal closure in $F$ of $\langle a^5,b^5,[a,b] \rangle$, so $F/H$ is elementary abelian of order $25$. Then the automorphism of $F/H$ that maps every element to its square does not lift to an automorphism of $F$.