Determine under what circumstances an involution on R is an automorphism

An automorphism of a ring is a homomorphism to itself which is bijective.

Since your $\alpha$ is already bijective, it is an automorphism if and only if it is a homomorphism.

Let me write down the conditions for being a homomorphism.

$$α(r_1 + r_2) = α(r_1) + α(r_2), α(r_1r_2) = α(r_1)α(r_2).$$ (Here I don't assume that the ring has a unit element, but it doesn't affect the argument.)

Do you notice the difference from the conditions of being involution?

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The only extra condition is $α(r_1r_2) = α(r_1)α(r_2)$. Thus if $\alpha$ is an involution and also an automorphism, then we must have $α(r_1)α(r_2)=α(r_1r_2) = α(r_2)α(r_1)$.

This being true for all $r_1,r_2$, we may substitute $r_1=α(x)$ and $r_2=α(y)$ for arbitrary $x,y$, and get $xy=yx$.

Therefore we have shown that, if there is an involution which is an automorphism, then the ring must be commutative.

Conversely, if the ring is commutative, then for any involution $\alpha$, we have $α(r_1r_2) = α(r_2)α(r_1)=α(r_1)α(r_2)$, which implies that it's an automorphism.

Since an involution is an antihomomorphism, and an automorphism is a homomorphism, we need both. That's equivalent to $\mathcal R$ being commutative.

You already showed bijectiveness.