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More generally, the following is Fact 2.5 of my paper with Michael Lieberman and Jiří Rosický on internal sizesLieberman, Rosický, and Vasey - Internal sizes in $\mu$-abstract elementary classes:

Theorem. Assume $\lambda$ and $\mu$ are regular cardinals and $2^{<\lambda} < \mu$. Then $\lambda \triangleleft \mu$ if and only if $\lambda \ll \mu$.

Assuming GCH, $2^{<\lambda} = \lambda$, so we recover Gabe's answer (the proof is the same).

More generally, the following is Fact 2.5 of my paper with Michael Lieberman and Jiří Rosický on internal sizes:

Theorem. Assume $\lambda$ and $\mu$ are regular cardinals and $2^{<\lambda} < \mu$. Then $\lambda \triangleleft \mu$ if and only if $\lambda \ll \mu$.

Assuming GCH, $2^{<\lambda} = \lambda$, so we recover Gabe's answer (the proof is the same).

More generally, the following is Fact 2.5 of Lieberman, Rosický, and Vasey - Internal sizes in $\mu$-abstract elementary classes:

Theorem. Assume $\lambda$ and $\mu$ are regular cardinals and $2^{<\lambda} < \mu$. Then $\lambda \triangleleft \mu$ if and only if $\lambda \ll \mu$.

Assuming GCH, $2^{<\lambda} = \lambda$, so we recover Gabe's answer (the proof is the same).

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More generally, the following is Fact 2.5 of my paper with Michael Lieberman and Jiří Rosický on internal sizes:

Theorem. Assume $\lambda$ and $\mu$ are regular cardinals and $2^{<\lambda} < \mu$. Then $\lambda \triangleleft \mu$ if and only if $\lambda \ll \mu$.

Assuming GCH, $2^{<\lambda} = \lambda$, so we recover Gabe's answer (the proof is the same).