If $\phi: X\to Y$ is a bijection, is $\bar\phi : S^Y\to S^X$ an isomorphism? 
Consider sets $X,Y \neq \emptyset$. $(S,\cdot)$ is a binary structure, i.e. $\cdot$ is the binary operation such that $\forall s_1,s_2\in S$, $s_1\cdot s_2 \in S$. $S^Y$ is the set of all functions from $Y$ to $S$. Similarly for $S^X$. Consider $$\bar\phi:S^Y \to S^X$$ such that for all $f:Y\to S$, $$\bar\phi(f)(x) = f(\phi(x)), \forall x\in X$$
Note that $S^X$ is also a binary structure (similarly for $S^Y$) with the operator $\otimes$, such that $$(f\otimes g)(x) = f(x)\cdot g(x), \forall x\in X$$
Prove that $\bar\phi$ is a homomorphism, i.e.
$$\bar\phi(f\otimes g)(x) = (\bar\phi(f)\otimes \bar\phi(g))(x), \forall x\in X$$

$$\bar\phi(f\otimes g)(x)  = (f\otimes g)(\phi(x)) = f(\phi(x))\cdot g(\phi(x)) = \bar\phi(f(x)) \cdot \bar\phi(g(x)) = (\bar\phi(f)\otimes\bar\phi(g))(x), \forall x\in X$$
I think this shows that this is a homomorphism. Is my proof fine?

Question: If $\phi: X\to Y$ is a bijection, is $\bar\phi : S^Y\to S^X$ an isomorphism? If yes, is this condition necessary for $\bar\phi$ to be a bijection, or will something weaker (such as injectivity of $\phi$) suffice?

Alright, let's say that $\phi$ is a bijection. Now I'll check if $\bar\phi$ is one-one. Assume $\bar\phi(f(x)) = \bar\phi(g(x)), \forall x\in X$ for some $f,g \in S^Y$. So, $f(\phi(x)) = g(\phi(x)), \forall x\in X$. I'm stuck here, what can I do next?
 A: Your proof that $\bar{\phi}$ is a homomorphism is fine.
For the proof of $\bar{\phi}$ being one-to-one note that since $\phi$ is surjective, the fact that $f(\phi(x))=g(\phi(x))$ for all $x\in X$ implies that $f(s)=g(s)$ for all $s\in S$, so necessarily $f=g$ as maps $X\to S$.
For surjectivity of $\bar{\phi}$, let $g\in S^X$ be a map $X\to S$. Define $f\in S^Y$ by $g(\phi^{-1}(y))$. This is well defined because $\phi$ is a bijection. Now, $\bar{\phi}(f)(x)=f(\phi(x))=g(\phi^{-1}(\phi(x)))=g(x)$, so $\bar{\phi}(f)=g$.
So in conclusion, $\bar{\phi}$ is an isomorphism.
A: $\bar\phi$ is one-to-one if $\phi$ is onto
To prove that $\bar\phi$ is one-to-one, we have to prove that $\bar\phi(f) = \bar\phi(g)$ implies $f=g$. And for this it is enough to have $f(y)=g(y)$ for all $y \in Y$.
$h$ is onto. Therefore for all $y \in Y$ it exists $x \in X$ such that $y = \phi(x)$ and
$$f(y) = f(\phi(x)) = \bar\phi(f(x)) = \bar\phi(f)(x)=\bar\phi(g(x))=g(\phi(x))=g(y)$$
proving that $\bar\phi$ is one-to-one as desired.
