Unrolling multi-variable mu (μ) expressions in type theory
I can't image they end up equivalent.


That's where you're wrong! In fact, starting with two different unfoldings of some expression you can always keep unfolding them in such a way that they "meet" again.

This property is formally called confluence.

More formally: define

$$ T \rightarrow_\tau T'$$

If $T'$ is the result of a repeated number of unfoldings of $T$. We have the following Theorem:

Suppose $$ T_1\ {}_\tau\!\leftarrow T\rightarrow_\tau T_2$$

There exists $T_3$ such that $$ T_1\rightarrow_\tau T_3\ {}_\tau\!\leftarrow T_2$$

The proof is a bit subtle. I'll just try and convince you using your own example: $$T = \mu\alpha\beta. 1 + \alpha\times\beta$$

The reductions you outlined contain a slight mistake, the correct "$\alpha$-reduction" is:

$$T\rightarrow_\tau \mu\beta.1 + T\times \beta\qquad (1)$$

and not $T\rightarrow_\tau \mu\alpha\beta.1 + (\mu\beta.1+\alpha\times\beta)\times\beta$ As you suggest.

The $\beta$-unfolding is:

$$T\rightarrow_\tau \mu\alpha.1+\alpha\times(\mu\beta.1+\alpha\times\beta)\qquad (2)$$

Let's show both of these lead to the same term. Take $U(\alpha)=\mu\beta.1+\alpha\times\beta$. We have
$$ T\rightarrow_\tau \mu\alpha.1+\alpha\times U(\alpha)$$
By the above reduction. Set $T_1=\mu\alpha.1 +\alpha\times U(\alpha)$. On the other hand
$$T\rightarrow_\tau 1+ T\times U(T)\qquad (3)$$

by composing (1) with a $\beta$-unfolding. Composing the second reduction with an $\alpha$-reduction gives
$$ 1 + T_1\times U(T_1)$$

But since $T\rightarrow_\tau T_1$, we can replace $T$ by $T_1$ in (3) to get identical terms.