is ∀x∃y(Lxy) and ∃x∀yLyx provable from each other?

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The proof I have is :

$\begin{aligned}1&\ &\forall x\exists yLxy\\ 2&\ &\exists y\ \text{(Lay)}\ \forall E_1\\ 3&\ &\text{Assume Lab}\\ 4&\ & \forall y Lyb\ \forall I_3∀yLyb\\ 5&\ &\exists x\forall yLyx \exists I_4\\ 6&\ & \exists x\forall yLyx ∃x∀yLyx\ \exists E_2 ∃, 3–5 \\\end{aligned}$

But I don't know if steps 4 and 5 are proper application of the universal and existential introductions.

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It is not true. Let $Lxy$ be $x=y$ and the universe be $\{1,2\}$ Then $\forall x \exists y Lxy$ is true because for whatever $x$ you pick you can make $y$ be the same. $\exists x \forall y Lyx$ is false because whatever $x$ you pick there is an unequal $y$.