The intuitive approach to conjugate priors is to try to deduce a family of distributions from the likelihood function. In the normal case, the likelihood is
\begin{align*}\ell(\mu,\Sigma|x_1,\ldots,x_n)&\propto|\Sigma|^{-n/2}\,\exp\left\{
-\frac{1}{2}\sum_{i=1}^n (x_i-\mu)^\text{T}\Sigma^{-1}(x_i-\mu)\right\}\\
&\propto|\Sigma|^{-n/2}\,\exp\left\{-\frac{1}{2}\sum_{i=1}^n (x_i-\bar{x})^\text{T}\Sigma^{-1}(x_i-\bar{x})\right\}\\&\qquad \times \exp\left\{-\frac{n}{2}(\bar{x}-\mu)^\text{T}\Sigma^{-1}(\bar{x}-\mu)\right\}\\
&\propto |\Sigma|^{-n/2}\,\exp\left\{-\frac{1}{2}\text{tr}\left(\Sigma^{-1}S_n \right)-\frac{n}{2}(\bar{x}-\mu)^\text{T}\Sigma^{-1}(\bar{x}-\mu)\right\}\\
\end{align*}where $$S_n=\sum_{i=1}^n (x_i-\bar{x})(x_i-\bar{x})^\text{T}$$So we have three items in this likelihood:
- a power of $|\Sigma|$;
- an exponential of a trace of $\Sigma^{-1}$ times another matrix;
- an exponential of a quadratic in $\mu$ with matrix $\Sigma^{-1}$.
And all three terms are stable by multiplication, i.e.
- $|\Sigma|^a\times |\Sigma|^b = |\Sigma|^{a+b}$;
- $\exp\left\{-\text{tr}\left(\Sigma^{-1}A\right)\right\}\times\exp\left\{-\text{tr}\left(\Sigma^{-1}B\right)\right\}=\exp\left\{-\text{tr}\left(\Sigma^{-1}[A+B]\right)\right\}$;
- $\exp\left\{-(a-\mu)^\text{T}\alpha\Sigma^{-1}(a-\mu)\right\}\times\exp\left\{-(b-\mu)^\text{T}\beta\Sigma^{-1}(b-\mu)\right\}$ remains an exponential of a quadratic in $\mu$ with matrix $\Sigma^{-1}$ (with an extra term of the form $\exp\left\{-\text{tr}\left(\Sigma^{-1}A\right)\right\}$ as this is not a perfect quadratic term).
This means that the likelihood induces a shape of prior that remains stable by multiplication with another term with this shape. Which is a way of defining conjugacy. So, if I take my prior to be
$$\pi(\mu,\Sigma)\propto|\Sigma|^{-\gamma/2}\,\exp\left\{-\frac{1}{2}\text{tr}\left(\Sigma^{-1}\Xi \right)-\frac{\nu}{2}(\xi-\mu)^\text{T}\Sigma^{-1}(\xi-\mu)\right\}$$the posterior will look the same, except that $\gamma,\Xi,\nu,\xi$ will change.