Are matrix $p$-norms unitary invariant?

Consider a matrix $X \in \mathbb{R}^{N \times N}$. Let $\| X \|_p$ be its $p$-norm

$$\| X\|_p = \left( \sum_{ij} |X_{ij}|^p \right)^{\frac 1p}$$

Is $\|X\|_p$ unitary invariant? That is, given any unitary matrix $U$, $V$, is $\|X\|_p$ = $\|UXV\|_p$?

My strategy is to formulate the $p$-norm into some kind of singular value or trace, but I failed to do this. If it is not unitary invariant, how can I prove it?

I suppose you already know about $p=2$ which is the frobenius norm of the matrix and is the only norm which is unitarily invariant among all $p$. To prove it is unitarily invariant, try expressing the frobenius norm in terms of trace operator and then use the circularity of trace. To prove other cases are not unitarily invariant, try picking examples for any given $p$. I don't know how to prove it. I am checking on it.

Update: First of all, it is enough to prove that vector $p-$ norms are not unitarily invariant. This follows from the factor that matrix $p-$ norms can be expressed as vector $p-$norms of its columns. The idea is to prove that for any given $p\neq 2$, you can find a vector and a unitary matrix such that unitary invariance doesn't hold. Take $x=[1,0,\dots,0]^T$. Note that $(||x||_p)^p=1$. Thus, we need to find a unitary matrix $U$ such that $(||Ux||_p)^p\neq 1$. Consider a unitary $U$ whose first column is given by $c_1=\left[\frac{1}{\sqrt{N}},\frac{1}{\sqrt{N}},\dots,\frac{1}{\sqrt{N}}\right]$ (eg: DFT Matrix). Now, note that $(||c_1||_p)^p=N^{1-\frac{p}{2}}$(note the case $p=2$). Now observe that $(||Ux||_p)^p=(||c_1||_p)^p\neq 1 = (||x||_p)^p$. Thus for every $p$ except $p=2$, you can construct a counter example.