Isomorphism?

Instead of φ(x) I use f(x)

Let r > 0 and b=+sqrt(r)
Then f(r) = f(b*b) = f(b)*f(b) = f^2(b) > 0 since it is a square.
So if r>0 then f(r) > 0

Also f(r)*f(r) = f(r^2) = f((-r)*(-r)) = (f(-r))^2
And we have f^2(r) = f^2(-r)
so that either f(-r) = f(r) or f(-r) = -f(r)
If f(-r) = f(r) then f() is not bijective ( since f(r) and f(-r) map to the same number) contrary to what we are given.
Hence f(-r) = -f(r) < 0 since f(r) > 0 when r > 0

an isomorphism φ : R* → R* is x^4/x

One simple instance for φ : R* → R* to be a group isomorphism
is to define φ(x) = 1/x.

Then, φ is a group homomorphism, because for any nonzero a, b in R:
φ(ab) = 1/(ab) = (1/a) (1/b) = φ(a) φ(b).

Next, φ is onto, because given a nonzero x in R, 1/x is also a nonzero real number, and φ(1/x) = 1/(1/x) = x, as required.

Finally, φ is 1-1, because ker φ = {1}:
ker φ = {x in R* : φ(x) = 1/x = 1} = {1}.

Hence, φ is a group isomorphism.

Note: More generally, if we fix any odd integer k,
φ : R* → R* defined by φ(x) = x^k gives an isomorphism.

I hope this helps!

φ(x) = x^3 is one such mapping that is not equal to the identity map.

Im making the assumption, of course, that the operation of consideration is multiplication. Note that it doesnt have to be multiplication. Identifying a group by its set and without specifying an operation is poor practice, but I frequently see this. Its presumed that if you dont specify an operation then it is the most trivial of operations that satisfy. I just think this is an intuition-based assumption-making practice that utterly lacks rigor. It frustrates me as a pure mathematician and who else but pure mathematicians would study abstract algebra?

Proving the statement, though, a bit more work. Give me a minute.

I think a good proof would be one from contradiction.

We know that as a homomorphism the rule φ(ab) = φ(a)φ(b) holds. We can also deduce that φ((-a)(-b)) = φ(ab), which = φ(a)φ(b), by definition. The negative sign could be manipulated like that on the left side because multiplication is commutative and associative (one good reason to assume multiplication is the operation).

Let as assume for contradictive hypothesis that also φ((-a)b) = φ(a(-b)) = φ(a)φ(b) = φ(ab).

This means that ab and -ab both map to φ(ab). φ is therefore not bijective, which is a contradiction. This means then that in fact φ(-ab) has to map to φ(-a)φ(b), or equivalently to φ(a)φ(-b), where φ(a) not equal to φ(-a) and φ(b) not equal to φ(-b).

This so far doesnt exactly prove your conditionals, but its close.