Isomorphism of maximal subgroups fix the Frattini?

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Let $G$ be a finite $p$-group and $M$ and $N$ be two maximal subgroups of $G$. If $f: M \to N$ is an isomorphism, is it true that $f(\Phi(G)) \subseteq \Phi(G)$, where $\Phi(G)$ denotes the Frattini subgroup of $G$?

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No. Let $G$ be the dihedral group of order 8, and $M = N$ a Klein four-subgroup. The Frattini subgroup of $G$ is the order-2 subgroup generated by a 180-degree rotation, and this is not fixed by all automorphisms of $M$.