Is $\operatorname{Aut}(C_2\times C_4)=C_2$?

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What is the automorphism group of $G:=C_2\times C_4$?

Thoughts . . .

The only nontrivial automorphism I can think of is, for $G=\langle a, b\mid a^2, b^4, ab=ba\rangle$, the one that swaps the roles of $b$ and $b^3$, acting as an identity on $a$.

How do I show that this is the only one?

Please help :)

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The direct approach (which isn't too bad in this case): A homomorphism (and by extension an automorphism) is uniquely determined by where it sends generators of the domain group. Pick a set of generators (like your $a$ and $b$) and go through all possible homomorphisms systematically by where they send the generators, and check whether they are automorphisms.

Tip: Excluding a lot of possibilities at once is allowed. For instance, if $f:G\to G$ is an automorphism, then the orders of $f(a)$ and $f(b)$ must be the same as the orders of $a$ and $b$. Thus you can clearly exclude all $f$ that have $f(a) = e$ or $f(b) = e$, regardless of what happens to the other generator.