How to get the inverse of Kronecker self-product (Kronecker square root)

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I am basically trying to solve for $M \in \mathbb{R}^{m \times n}$ below, such that for any vector $\mathbf{x} \in \mathbb{R}^{n^2}$ and some fixed $C \in \mathbb{R}^{m^2 \times n^2}$:

$$ (M \otimes M) \mathbf{x} = C \mathbf{x}$$

If there is a unique solution, I guess this is the same as finding $M$ such that $(M \otimes M) = C$.

Are there conditions under which there are solutions? Any knowledge about this problem would be useful, as I'm not too familiar with it; does it have a name? Are there certain fields where solving for M like this comes up?

Edit: first version of the question incorrectly stated C was square and symmetric.