Using induction on a word-concatenating function 
Let $\rho = \{ \cdot \}$ be a signature with a binary functional symbol. We define the $\rho$-structure  $\mathcal{S} := ( S , \cdot^{\mathcal{S}} )$, where $S$ is the set of finite words over a finite alphabet $\Sigma$, such that $S = \Sigma^*$, and $\cdot^{\mathcal{S}}$ the normal concatenatination of words, $x \cdot^{\mathcal{S}} y = xy \in S$, for all $x,y \in S$. Also, let $h: \Sigma \rightarrow \Sigma$ be any bijection.
Show using induction, that the function $f: S \rightarrow S$ is an automorphism:

$$f(w) :=\begin{cases}
       h(a_1) \cdot^{\mathcal{S}} f(a_2 \ldots a_n) & \text{if } w = a_1a_2 \ldots a_n , \text{ with } a_i \in \Sigma , \text{ for all } i \in \{ 1, \ldots , n \} , \\
       \lambda & \text{if } w = \lambda .
    \end{cases}$$

Where $\lambda$ is the empty word.

I'm practically lost and don't know where to start. What I understand:

*

*Since $f$  has to be a automorphism, it also must be an endomorphism and an isomorphism. Being an endomorphism is given, since it maps $S$ to itself.

*To show that $f$ is an isomorphism, we need to show that it is injective and surjective.

I'm not sure how to do this with induction and wasn't even able to come up with the induction hypothesis. Since we're trying to prove a property of a function, how could we use $P(n)$ and $P(n+1)$ (in this case $P(w)$ and $P(w \cdot v)$)? I'm pretty sure I'm looking at this the wrong way and would really appreciate any hints.
 A: Welcome to MSE!
In order to work with induction, we need something to induct on. You've correctly identified that one way to do this might be to look at the length of your words. After all, you can think of a word as being either empty (of length $0$) or a shorter word with a new character at the end (so it has length $n+1$). If you're familiar with structural induction we can actually use this to induct directly on the word itself, but for now let's work with the length.
Ok. So we want to show that $f : \Sigma^* \to \Sigma^*$ as defined in your problem is an isomorphism. You're correct that this comes down to showing it is an endomorphism and a bijection, but you've missed an important detail about endomorphism-ness. It's not enough to check that it maps $\Sigma^*$ to $\Sigma^*$! We also need to check that it's a homomorphism. That is, we need to check that $f(u \cdot v) = f(u) \cdot f(v)$.

So, how do we verify these properties? Well, let's see if there are any particularly easy cases to verify... We've been told to think about induction, so why not try it with words of length $0$? That seems like a reasonable base case to me.
Since $\lambda$ is the unique word of length $0$, we want to show that

*

*(homomorphism, base case) $f(\lambda \cdot w) = f(\lambda) \cdot f(w)$ for any word $w$

*(bijection, base case) $f \upharpoonright_{\Sigma^0} : \Sigma^0 \to \Sigma^0$ is a bijection

Here we've written $\Sigma^0$ for the words of length $0$. More generally, to check that $f$ is a bijection on all of $\Sigma^*$, it will be enough to show that $f$ is a bijection on our words of length $n$ for each $n$ (do you see why?).
So, let's verify these claims. These are each a simple computation:
$$
f(\lambda \cdot w) \overset{(1)}{=} 
f(w) \overset{(2)}{=} 
\lambda \cdot f(w) \overset{(3)}{=}
f(\lambda) \cdot f(w)
$$
In step $(1)$ we used the definition of $\cdot$. In step $(2)$ we used the definition of $\cdot$ again (but on the outside this time). In step $(3)$ we use the definition of $f$.
Next we want to check bijectivity on words of length $0$. But since $\lambda$ is the only word of length $0$, and we know that $f(\lambda) = \lambda$ we're done! It's a fairly simple bijection, but it's a bijection nonetheless.

Now it's time for the inductive step. We can now assume (as our Inductive Hypothesis) that

*

*(homomorphism, IH) for words $u$ of length $n$, $f(u \cdot w) = f(u) \cdot f(w)$ for all words $w$.

*(bijection, IH) $f \upharpoonright_{\Sigma^n} : \Sigma^n \to \Sigma^n$ is a bijection on words of length $n$.

We want to show that these results continue to hold for words of length $n+1$.
Let's prove bijection together. I'll leave homomorphism to you.
To show $f \upharpoonright_{\Sigma^{n+1}}$ is a bijection, we want to show it is injective and surjective. We might as well start with injective.
Let $au$ and $bv$ be two words of length $n+1$. Here $a, b \in \Sigma$ are single letters, and $u, v \in \Sigma^n$ are the rest of the word. To show injectivity, we want to show that if $f(au) = f(bv)$ then actually $au = bv$.
Now if $f(au) = f(bv)$, what must we know? Well:

*

*$f(au) = h(a) \cdot f(u)$ by the definition of $f$

*$f(bv) = h(b) \cdot f(v)$ for the same reason

So $h(a) \cdot f(u) = h(b) \cdot f(v)$. What does this buy us? Well, if two words are the same, then they had better have the same first letter! So $h(a) = h(b)$.
But since $h$ is a bijection, this says $a=b$!
Moreover, if the two words are the same, then the rest of the characters had better be the same too! So $f(u) = f(v)$, and now by applying our inductive hypothesis, we see that $u=v$.
So $a=b$ and $u=v$... Now we're done! That means $au=bv$, and so $f$ is injective on words of length $n+1$.
Surjectivity is similar, so I'll go a bit more quickly through it. Let's let $bv$ be my favorite word of length $n+1$. We want to show it's in the image of $f$.
Well, we know $h$ is a bijection, so we can find a letter $a$ with $h(a) = b$. Similarly, by induction, we know that $f$ is a bijection on words of length $n$, so we can find a word $u$ with $f(u) = v$.
Putting these pieces together gives $f(au) = bv$ (do you see why?) which shows $f$ is also surjective on words of length $n+1$.
Can you use these same ideas to show that $f$ is a homomorphism?

I hope this helps ^_^
