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Now we want to specialize to the case $X = RP^n,$ for some $n$. However it turns out that it's much easier to classify maps not into $RP^n$ itself, but rather into $RP^\infty.$ Namely, maps from some space $X$ to $RP^\infty$ up to homotopy are always classified by $H^1(X, \mathbb{Z}/2)$ (equivalently, this is group homomorphisms from $\pi_1(X)$ into $\mathbb{Z}/2$). Now the difference between $RP^n$ and $RP^\infty$ isn't actually too terrible. Namely, the CW approximation theorem tells us that any map $RP^n\to RP^\infty$ actually can be chosen up to homotopy such that it factorizes through $RP^n\subset RP^\infty.$ So we can compute $$\pi_0\mathrm{Maps}(RP^n, RP^\infty) \cong H^1(RP^n, \mathbb{Z}/2)\cong \mathbb{Z}/2.$$ The element $0\in \mathbb{Z}/2$ corresponds to the trivial map $$*:RP^n\to RP^n$$ mapping everything to a basepoint in $RP^\infty$ and the nontrivial element $1\in \mathbb{Z}/2$ corresponds to the identity map, $$id:RP^n\to RP^n\subset RP^\infty.$$ And any other map $RP^n\to RP^n$ will be homotopy equivalent to one of these as an element of $\mathrm{Maps}(RP^n, RP^\infty)$. However there's a catch: while any map $RP^n\to RP^n$ is homotopy equivalent to one of the maps you wrote down as a map to $RP^\infty$, the homotopy between the two maps might not live in $RP^n$. So a priori, there can be multiple homotopy classes of self-maps of $RP^n$ homotopic to one of these maps in $RP^\infty$ (and indeed, sometimes there are). To get a handle on how badly maps to $RP^\infty$ are undercounting, you can apply the CW approximation theorem again to see that any homotopy between two maps $RP^n\to RP^\infty$ will factorize, up to homotopy, through $$RP^{n+1}\subset RP^\infty.$$ The "error" of such a homotopy existing in $RP^n$ will be classified by a map to the quotient, $$RP^n\times [0,1]\to RP^{n+1}/RP^n\cong S^{n+1},$$ taking both $RP^n\times \{0\}, RP^n\times \{1\}$ to the basepoint, in other words, a based map from the space $$RP^n_+\wedge S^1 = \Sigma(RP^n_+)$$ (you can think of this as the ordinary suspension $\Sigma(RP^n)$ with the two suspension points identified) to $S^{n+1}.$ Let's write $$D: = \pi_0\text{Maps}(RP^n_+\wedge S^1, S^n)$$ for the set of possible such "defects" of a homotopy in $RP^{n+1}$ restricting to $RP^n$.

Thus for each of the maps $*, id: RP^n\to RP^n,$ there can be at worst $D$ worth of distinct other homotopy classes of self-maps $RP^n\to RP^n$ isomorphic to it as maps to $CP^{\infty}.$ Since a map homotopic to $*$ cannot be an automorphism (it would have to induce the trivial map on $H^1$ which cannot come from an automorphism), we can restrict our attention to maps homotopic in $CP^\infty$ to $id:RP^n\to RP^n.$ A priori, there could have been elements of $D$ which are not realizable as "defects" of homotopies between maps $RP^n\to RP^n$, but in this case we don't run into this problem: indeed, every element of $D$ occurs as the defect of some homotopy $RP^n\times [0,1]\to RP^{n+1}$ between the identity $id:RP^n\to RP^n$ and another map. Namely, recall that we have realized $D \cong H^n(RP^n)$ as a cyclic group, either $\mathbb{Z}$ or $\mathbb{Z}/2$ (depending on parity). Let $\alpha$ be a generator of this group. Then every element of $D$ can be written $k\alpha$ for some $k\in \mathbb{Z}$. By doing a calculation, you can see that each element $k\alpha\in D$ is realized as the defect of the homotopy $$[0,1]\cdot RP^n\to RP^{n+1}$$ induced by the map $[0,1]\times S^n\to S^{n+1}$ given by rotating $S^n$ in a circle around some ($n-1$-dimensional) axis inside $S^{n+1},$ by an angle of $$k\cdot \pi.$$ Now if $n$ is even, we see the resulting "new" map $RP^n\to RP^n$ is once again the identity. If $n$ is odd, the new map is induced from the "reflection" map given by $$\sigma:(x_1,x_2,\dots, x_n)\mapsto (-x_1,x_2, \dots, x_n)$$ (in some coordinates). Thus from what we've done so far, there can be at most two homotopy invertible self-maps up to homotopy $$RP^n\to RP^n$$ for any $n$, namely $id$ and $\sigma.$ It remains to check whether the induced two self-maps $RP^n\to RP^n$ are homotopic to each other. When $n$ is odd, they cannot be homotopic to each other since $RP^n$ is orientable, and $\sigma$ reverses orientation (so $\sigma$ can be distinguished from $id$ by looking at action on $H^n$). But when $n$ is even, the map $\sigma$ is homotopy equivalent via sphere rotations to the map $(x_1,x_2\dots, x_n)\mapsto (-x_1,-x_2\dots, -x_n),$ and this induces a homotopy between $\sigma$ and $id$ as maps $RP^n\to RP^n.$ Thus we have $$\pi_0(\mathrm{Aut}(RP^n)) \cong \begin{cases} \{id\}, & n \text{ even}\\ \{id, \sigma\}, & n \text{ odd}. \end{cases} $$ As mentioned, $RP^n$-bundles on $S^1$ are classified by the same data, so your guess is correct.

Now we want to specialize to the case $X = RP^n,$ for some $n$. However it turns out that it's much easier to classify maps not into $RP^n$ itself, but rather into $RP^\infty.$ Namely, maps from some space $X$ to $RP^\infty$ up to homotopy are always classified by $H^1(X, \mathbb{Z}/2)$ (equivalently, this is group homomorphisms from $\pi_1(X)$ into $\mathbb{Z}/2$). Now the difference between $RP^n$ and $RP^\infty$ isn't actually too terrible. Namely, the CW approximation theorem tells us that any map $RP^n\to RP^\infty$ actually can be chosen up to homotopy such that it factorizes through $RP^n\subset RP^\infty.$ So we can compute $$\pi_0\mathrm{Maps}(RP^n, RP^\infty) \cong H^1(RP^n, \mathbb{Z}/2)\cong \mathbb{Z}/2.$$ The element $0\in \mathbb{Z}/2$ corresponds to the trivial map $$*:RP^n\to RP^n$$ mapping everything to a basepoint in $RP^\infty$ and the nontrivial element $1\in \mathbb{Z}/2$ corresponds to the identity map, $$id:RP^n\to RP^n\subset RP^\infty.$$ And any other map $RP^n\to RP^n$ will be homotopy equivalent to one of these as an element of $\mathrm{Maps}(RP^n, RP^\infty)$. However there's a catch: while any map $RP^n\to RP^n$ is homotopy equivalent to one of the maps $id, *$ as a map to $RP^\infty$, the homotopy between the two maps might not live in $RP^n$. So a priori, there can be multiple homotopy classes of self-maps of $RP^n$ homotopic to one of these maps in $RP^\infty$ (and indeed, sometimes there are). To get a handle on how badly maps to $RP^\infty$ are undercounting, you can apply the CW approximation theorem again to see that any homotopy between two maps $RP^n\to RP^\infty$ will factorize, up to homotopy, through $$RP^{n+1}\subset RP^\infty.$$ The "error" of such a homotopy existing in $RP^n$ will be classified by a map to the quotient, $$RP^n\times [0,1]\to RP^{n+1}/RP^n\cong S^{n+1},$$ taking both $RP^n\times \{0\}, RP^n\times \{1\}$ to the basepoint, in other words, a based map from the space $$RP^n_+\wedge S^1 = \Sigma(RP^n_+)$$ (you can think of this as the ordinary suspension $\Sigma(RP^n)$ with the two suspension points identified) to $S^{n+1}.$ Let's write $$D: = \pi_0\text{Maps}(RP^n_+\wedge S^1, S^n)$$ for the set of possible such "defects" of a homotopy in $RP^{n+1}$ restricting to $RP^n$.

Thus for each of the maps $*, id: RP^n\to RP^n,$ there can be at worst $D$ worth of distinct other homotopy classes of self-maps $RP^n\to RP^n$ homotopic to it as maps to $CP^{\infty}.$ Since a map homotopic to $*$ cannot be an automorphism (it would have to induce the trivial map on $H^1$ which cannot come from an automorphism), we can restrict our attention to maps homotopic in $CP^\infty$ to $id:RP^n\to RP^n.$ A priori, there could have been elements of $D$ which are not realizable as "defects" of homotopies between maps $RP^n\to RP^n$, but in this case we don't run into this problem: indeed, every element of $D$ occurs as the defect of some homotopy $RP^n\times [0,1]\to RP^{n+1}$ between the identity $id:RP^n\to RP^n$ and another map. Namely, recall that we have realized $D \cong H^n(RP^n)$ as a cyclic group, either $\mathbb{Z}$ or $\mathbb{Z}/2$ (depending on parity). Let $\alpha$ be a generator of this group. Then every element of $D$ can be written $k\alpha$ for some $k\in \mathbb{Z}$. By doing a calculation, you can see that each element $k\alpha\in D$ is realized as the defect of the homotopy $$[0,1]\cdot RP^n\to RP^{n+1}$$ induced by the map $[0,1]\times S^n\to S^{n+1}$ given by rotating $S^n$ in a circle around some ($n-1$-dimensional) axis inside $S^{n+1},$ by an angle of $$k\cdot \pi.$$ Now if $n$ is even, we see the resulting "new" map $RP^n\to RP^n$ is once again the identity. If $n$ is odd, the new map is induced from the "reflection" map given by $$\sigma:(x_1,x_2,\dots, x_n)\mapsto (-x_1,x_2, \dots, x_n)$$ (in some coordinates). Thus from what we've done so far, there can be at most two homotopy invertible self-maps up to homotopy $$RP^n\to RP^n$$ for any $n$, namely $id$ and $\sigma.$ It remains to check whether the induced two self-maps $RP^n\to RP^n$ are homotopic to each other. When $n$ is odd, they cannot be homotopic to each other since $RP^n$ is orientable, and $\sigma$ reverses orientation (so $\sigma$ can be distinguished from $id$ by looking at action on $H^n$). But when $n$ is even, the map $\sigma$ is homotopy equivalent via sphere rotations to the map $(x_1,x_2\dots, x_n)\mapsto (-x_1,-x_2\dots, -x_n),$ and this induces a homotopy between $\sigma$ and $id$ as maps $RP^n\to RP^n.$ Thus we have $$\pi_0(\mathrm{Aut}(RP^n)) \cong \begin{cases} \{id\}, & n \text{ even}\\ \{id, \sigma\}, & n \text{ odd}. \end{cases} $$ As mentioned, $RP^n$-bundles on $S^1$ are classified by the same data, so your guess is correct.

Your answer is correct, but it's a little subtle. Here I should note that I'm interpreting your question as a purely topological one (ignoring smooth structure), and that by "bundle" you mean "Serre bundle". If you care about smooth bundles, see Tom Goodwillie's answer.

Your answer is correct if appropriately understood, but it's a little subtle. Here I should note that I'm interpreting your question as a purely homotopy theoretic one (in particular ignoring smooth structure), and that by "bundle" you mean "Serre bundle". If you care about smooth bundles, see Tom Goodwillie's answer.

Your answer is correct, but it's a little subtle.

Your answer is correct, but it's a little subtle. Here I should note that I'm interpreting your question as a purely topological one (ignoring smooth structure), and that by "bundle" you mean "Serre bundle". If you care about smooth bundles, see Tom Goodwillie's answer.