This is part $2$ of a $3$ part problem, I'll link the other parts once I've created questions for them and they've been resolved. I believe I have a proof for the problem but I'm a little unsure on how to tie a bow around the proof of surjectivity.
Problem Statement
Prove that for any subgroup $H <G$ and any element $g \in G$ that $c_{g}: H \to gHg^{-1}$ defined by $c_{g}(h) = ghg^{-1}$ is an isomorphism.
Relevant Proposition
Let $(G, \ast)$ be a group and $g,h,k \in G$. If $g \ast h = k \ast h$ then, $g = k$. Likewise, if $h \ast g = h \ast k$ then, $g = k$
Proof Attempt
Let $h,h' \in H$ then we have $$c_{g}(hh') = ghh'g^{-1} = gh(e)h'g^{-1} = gh(g^{-1}g)h'g^{-1} = (ghg^{-1})(gh'g^{-1}) = (c_{g}(h))(c_{g}(h'))$$ Thus $c_{g}$ is a homomorphism.
Now, assume $c_{g}(h) = ghg^{-1} = gh'g^{-1} = c_{g}(h')$ then by the proposition we have from left cancellation first: $$g(hg^{-1}) = g(h'g^{-1}) \implies hg^{-1} = h'g^{-1}$$ Now, by right cancellation: $$ hg^{-1} = h'g^{-1} \implies h = h'$$ Thus $c_g$ is injective. Now, since $c_g$ is injective and the order of $H$ is equivalent to the order of $gHg^{-1}$, by definition of $gHg^{-1}$, we can see that $c_g$ is surjective and hence bijective. Consequently $c_g$ is an isomorphism.
Questions/Concerns
First of all, is the proof, as a whole, sound? I'm pretty confident about the first $2$ parts, I just don't know if I'm providing enough justification that $\lvert H \rvert = \lvert gHg^{-1} \rvert$. To me, it makes sense that $\lvert H \rvert = \lvert gHg^{-1} \rvert$ from the definition of $gHg^{-1} = \{ ghg^{-1} \mid h \in H\}$ Since the conjugate of $H$ by $g$ is simply every element of $H$ multiplied on the right and left by the same $2$ elements of $G$. Should I mention this, or provide some other justification or is it fine to just say by the definition?
From there if every element in the range has at most $1$ element in the domain mapped to it (by injectivity) and both the domain and the range have the same number of elements then we can say that the map must be surjective, and hence bijective, correct? Since a function is a rule that assigns to each element of the domain exactly one element of the range.
Your proof of the homomorphism property and injectivity are good.
This is a very common mistake: $|H|=|gHg^{-1}|$ does not imply that the map is surjective. This is only true when the groups (or just sets) are finite. To show surjectivity just use the definition of surjectivity (it is very straightforward).
Just some food for thought: another way to show that $c_g$ is bijective is to construct an inverse map. Can you think of what the inverse of $c_g$ would be?
An example illustrating my above point is the following. The map $\mathbb Z\to\mathbb Z$ sending $x\mapsto 2x$ is injective, and clearly $|\mathbb Z|=|\mathbb Z|$, but the map is not surjective.