Are isomorphisms automorphisms? Suppose we have a set $X$ and two group operations on the set $\oplus,+$. If they are isomorphic, does it follow that $\oplus=+$?
I am unable to think of counter-examples, and this makes sense to me at an intuitive level because "isomorphism" means something in my head like "the operations are the same, just with a different set." However, I'm unable to prove it using the standard definition of $f(x\oplus y)=f(x)+f(y)$.
 A: No. Let $G$ be a nontrivial group under the operation $\bullet$ and let $f:G\to G$ be a permutation of its elements (which we don't regard as a homomorphism). Define $\circ$ on $G$ by $f(a\circ b)=f(a)\bullet f(b)$.
Then $(G,\circ)\cong(G,\bullet)$ but $\circ,\bullet$ needn't be the same operation. Essentially this means we can make the underlying set of $G$ into the group $G$ in many different ways by relabelling its elements.
A: Consider $G=\mathbb R$ to be the set of real numbers. Let $+: \mathbb R\times \mathbb R\rightarrow \mathbb R$ be the usual addition operation, and let $f:\mathbb R\rightarrow \mathbb R$ be the bijective function defined by $f(a):=a^3$. Now, define $\oplus: \mathbb R\times \mathbb R\rightarrow \mathbb R$ to be the operation defined by $$a\oplus b:=f^{-1}(f(a)+f(b))=(a^3+b^3)^{1/3}.$$ You can check that $(\mathbb R,+)$ and $(\mathbb R,\oplus)$ are both groups, moreover $f:(\mathbb R,\oplus)\rightarrow (\mathbb R,+)$ is automatically an isomorphism of groups; but $+$ and $\oplus$ are different as operations. 
