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I am just wondering if there is any connection between $\|A\|_2$ and $\|A^H\|_2$ where $A^H$ is the conjugate transpose of $A$ and $\| \|_2$ is the spectral norm. This problem is basically asking, like, if $A^HA$ and $AA^H$ has the same maximum eigenvalue.

PS: I found from Eigen of $AA^H$ and of $A^HA$ that $0$ might just be the eigenvalue of one of them, but this does not say about the maximum eigenvalue.

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    In fact all of the nonzero eigenvalues of $A^H A$ and $A A^H$ are the same; they are the squares of the singular values of $A$. More generally, if $A, B$ are two matrices such that the products $AB$ and $BA$ make sense, then all of the nonzero eigenvalues of $AB$ and $BA$ are the same. – 2017-02-24

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One can show that $\|A\|_2=\|A^H\|_2$, with a fairly easy proof coming from the $C^*$-identity: $\|A^HA\|_2=\|A\|_2^2$. From this, we have $$\|A\|_2^2=\|A^HA\|_2\leq\|A^H\|_2\|A\|_2$$ and $$\|A^H\|_2^2=\|AA^H\|_2\leq\|A^H\|_2\|A\|_2$$ which gives us $\|A\|_2\leq\|A^H\|_2\leq\|A\|_2$, hence equality.

You can also prove it via the result you mention: if $\sigma(A^HA)\cup\{0\}=\sigma(AA^H)\cup\{0\}$, then $$\|A\|_2^2=\max\{|\lambda|:\lambda\in\sigma(A^HA)\} =\max\{|\lambda|:\lambda\in\sigma(AA^H)\}=\|A^H\|_2^2.$$.